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Trigonometric Identities: Class 10 Maths Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 10 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based4 marks

A civil engineer is designing a ramp for a building. While checking structural calculations, she uses trigonometric expressions to model stress ratios and load distribution at a given angle θ.

A civil engineer is designing a ramp for a building. While checking the structural calculations, she writes the following trigonometric expression that models the stress ratio at angle θ:

(sin θ + cos θ)² + (sin θ − cos θ)²

She also needs to verify a second identity used in load-distribution calculations:

(tan²θ − sin²θ) = tan²θ · sin²θ

(i) Evaluate the engineer's stress-ratio expression: (sin θ + cos θ)² + (sin θ − cos θ)² [1 mark]
(ii) Using the identity sin²θ + cos²θ = 1, prove the load-distribution identity:
(tan²θ − sin²θ) = tan²θ · sin²θ [3 marks]

Show answer
(i) Evaluating (sin θ + cos θ)² + (sin θ − cos θ)²:

(sin θ + cos θ)² + (sin θ − cos θ)²

⟹ [sin²θ + 2 sin θ cos θ + cos²θ] + [sin²θ − 2 sin θ cos θ + cos²θ]

⟹ [sin²θ + cos²θ + 2 sin θ cos θ] + [sin²θ + cos²θ − 2 sin θ cos θ]

Using the identity sin²θ + cos²θ = 1:

⟹ [1 + 2 sin θ cos θ] + [1 − 2 sin θ cos θ]

⟹ 1 + 1

∴ (sin θ + cos θ)² + (sin θ − cos θ)² = 2

(ii) To Prove: tan²θ − sin²θ = tan²θ · sin²θ

Taking the Left Hand Side (LHS):

LHS = tan²θ − sin²θ

⟹ sin²θ/cos²θ − sin²θ [∵ tan θ = sin θ/cos θ]

⟹ sin²θ · (1/cos²θ − 1) [taking sin²θ common]

⟹ sin²θ · ((1 − cos²θ)/cos²θ)

Using the identity sin²θ + cos²θ = 1, we have 1 − cos²θ = sin²θ:

⟹ sin²θ · (sin²θ/cos²θ)

⟹ sin²θ · tan²θ

⟹ tan²θ · sin²θ

= RHS. Hence proved.
Q2Case-based4 marks

A structural engineer is designing a triangular roof truss. The truss forms a right-angled triangle where one acute angle is θ. The stress ratio at Joint A is modelled by the expression E = (sin θ + cosec θ)² + (cos θ + sec θ)² − (tan²θ + cot²θ). The engineer needs to confirm this always gives a constant value for uniform stress distribution.

A structural engineer is designing a triangular roof truss for a building. The truss forms a right-angled triangle where one acute angle is θ. During load analysis, the engineer uses trigonometric expressions to model stress ratios at various joints. The stress ratio at Joint A is given by the expression:

E = (sin θ + cosec θ)² + (cos θ + sec θ)² − (tan²θ + cot²θ)

The engineer needs to verify that this expression always gives a constant value, regardless of the angle θ, to confirm uniform stress distribution across all designs.

Based on the above context, answer the following:

(i) Write the expanded form of (sin θ + cosec θ)². [1 mark]

(ii) Write the expanded form of (cos θ + sec θ)². [1 mark]

(iii) The engineer claims the value of E is always 7. Verify whether this claim is correct by simplifying the full expression E = (sin θ + cosec θ)² + (cos θ + sec θ)² − (tan²θ + cot²θ). Show all steps.
[2 marks]

OR

(iii) The engineer uses a different stress model at Joint B, given by:
F = (sin θ/(1 + cos θ)) + ((1 + cos θ)/sin θ)
Prove that F = 2 cosec θ. [2 marks]

Show answer
Answer:

(i) Expanded form of (sin θ + cosec θ)²:

(sin θ + cosec θ)² = sin²θ + 2·sin θ·cosec θ + cosec²θ

∵ sin θ · cosec θ = 1

∴ (sin θ + cosec θ)² = sin²θ + 2 + cosec²θ

(ii) Expanded form of (cos θ + sec θ)²:

(cos θ + sec θ)² = cos²θ + 2·cos θ·sec θ + sec²θ

∵ cos θ · sec θ = 1

∴ (cos θ + sec θ)² = cos²θ + 2 + sec²θ

(iii) Main option — Simplifying E:

E = (sin θ + cosec θ)² + (cos θ + sec θ)² − (tan²θ + cot²θ)

Using results from (i) and (ii):

E = (sin²θ + 2 + cosec²θ) + (cos²θ + 2 + sec²θ) − (tan²θ + cot²θ)

⟹ E = (sin²θ + cos²θ) + 4 + cosec²θ + sec²θ − tan²θ − cot²θ

Applying the identity sin²θ + cos²θ = 1:

⟹ E = 1 + 4 + cosec²θ + sec²θ − tan²θ − cot²θ

Applying the identities sec²θ − tan²θ = 1 and cosec²θ − cot²θ = 1:

⟹ E = 5 + (cosec²θ − cot²θ) + (sec²θ − tan²θ)

⟹ E = 5 + 1 + 1

⟹ E = 7

∴ E = 7, which is a constant value independent of θ.

Hence, the engineer's claim is correct. ✓

─────────────────────────
OR
─────────────────────────

(iii) Alternative option — Proving F = 2 cosec θ:

F = sin θ/(1 + cos θ) + (1 + cos θ)/sin θ

Taking LCM as sin θ(1 + cos θ):

F = [sin²θ + (1 + cos θ)²] / [sin θ(1 + cos θ)]

Expanding the numerator:

Numerator = sin²θ + 1 + 2cos θ + cos²θ

⟹ = (sin²θ + cos²θ) + 1 + 2cos θ

Applying the identity sin²θ + cos²θ = 1:

⟹ = 1 + 1 + 2cos θ

⟹ = 2 + 2cos θ

⟹ = 2(1 + cos θ)

∴ F = 2(1 + cos θ) / [sin θ(1 + cos θ)]

⟹ F = 2 / sin θ

⟹ F = 2 cosec θ

= RHS. Hence proved.
Q3Case-based4 marks

A structural engineer is designing a triangular roof truss. To verify the stability calculations, she uses trigonometric expressions involving the angle θ that the roof makes with the horizontal. She models two load-related expressions and one structural identity to confirm the design is mathematically consistent.

A structural engineer is designing a triangular roof truss. To verify the stability calculations, she uses the following trigonometric expression involving the angle θ that the roof makes with the horizontal:

(sin θ + cos θ)² + (sin θ − cos θ)²

She also needs to simplify a second expression for load distribution:

(1 − sin²θ)(1 + tan²θ)

(i) Evaluate the first expression and state its value. [1]
(ii) Simplify the second expression to its simplest form. [1]
(iii) The engineer uses the identity:

(cosec θ − sin θ)(sec θ − cos θ) = 1/(tan θ + cot θ)

Prove this identity. [2]

Show answer
(i) Evaluating (sin θ + cos θ)² + (sin θ − cos θ)²:

⟹ (sin²θ + 2 sin θ cos θ + cos²θ) + (sin²θ − 2 sin θ cos θ + cos²θ)

⟹ (sin²θ + cos²θ) + 2 sin θ cos θ + (sin²θ + cos²θ) − 2 sin θ cos θ

Using the identity sin²θ + cos²θ = 1:

⟹ 1 + 2 sin θ cos θ + 1 − 2 sin θ cos θ

⟹ 2

∴ The value of the expression is 2.

---

(ii) Simplifying (1 − sin²θ)(1 + tan²θ):

Using sin²θ + cos²θ = 1 ⟹ 1 − sin²θ = cos²θ

Using 1 + tan²θ = sec²θ

⟹ cos²θ × sec²θ

⟹ cos²θ × 1/cos²θ

⟹ 1

∴ The simplified value of the expression is 1.

---

(iii) To Prove: (cosec θ − sin θ)(sec θ − cos θ) = 1/(tan θ + cot θ)

Taking the LHS:

(cosec θ − sin θ)(sec θ − cos θ)

⟹ (1/sin θ − sin θ)(1/cos θ − cos θ)

⟹ [(1 − sin²θ)/sin θ] × [(1 − cos²θ)/cos θ]

Using 1 − sin²θ = cos²θ and 1 − cos²θ = sin²θ:

⟹ [cos²θ/sin θ] × [sin²θ/cos θ]

⟹ sin θ · cos θ ... (i)

Now taking the RHS:

1/(tan θ + cot θ)

⟹ 1/(sin θ/cos θ + cos θ/sin θ)

⟹ 1/[(sin²θ + cos²θ)/(sin θ cos θ)]

Using sin²θ + cos²θ = 1:

⟹ 1/[1/(sin θ cos θ)]

⟹ sin θ · cos θ ... (ii)

From (i) and (ii),

LHS = sin θ cos θ = RHS

= RHS. Hence proved.
Q4Case-based4 marks

A civil engineer is designing a ramp for a building entrance. She models the ramp as a right triangle and uses trigonometric ratios to check structural integrity. Her design is valid only if the expression (sin²θ + cos²θ) + tan²θ − sec²θ + cot²θ − cosec²θ equals 1.

A civil engineer is designing a ramp for a building entrance. To check the structural integrity, she models the ramp as a right triangle and uses trigonometric ratios. In her calculations, she arrives at the following expression that must equal 1 for the design to be valid:

(sin²θ + cos²θ) + tan²θ − sec²θ + cot²θ − cosec²θ

Based on this, answer the following parts:

(i) What is the value of sin²θ + cos²θ? [1 mark]
(ii) What is the value of sec²θ − tan²θ? [1 mark]
(iii) Evaluate the complete expression (sin²θ + cos²θ) + tan²θ − sec²θ + cot²θ − cosec²θ and verify whether the engineer's design condition (value = 1) is satisfied. [2 marks]

Show answer
(i) Using the Pythagorean trigonometric identity:

sin²θ + cos²θ = 1

∴ The value of sin²θ + cos²θ is 1.

(ii) Using the Pythagorean trigonometric identity:

1 + tan²θ = sec²θ

⟹ sec²θ − tan²θ = 1

∴ The value of sec²θ − tan²θ is 1.

(iii) The given expression is:

E = (sin²θ + cos²θ) + tan²θ − sec²θ + cot²θ − cosec²θ

Rearranging the terms:

E = (sin²θ + cos²θ) − (sec²θ − tan²θ) − (cosec²θ − cot²θ)

Using the three standard Pythagorean identities:

• sin²θ + cos²θ = 1
• sec²θ − tan²θ = 1 [since 1 + tan²θ = sec²θ]
• cosec²θ − cot²θ = 1 [since 1 + cot²θ = cosec²θ]

Substituting:

E = 1 − 1 − 1

⟹ E = −1

∴ The value of the expression is −1, which is NOT equal to 1.

Hence, the engineer's design condition is NOT satisfied, and she must recheck her calculations.
Q5Case-based4 marks

A civil engineer is designing a ramp for a warehouse loading dock. She uses trigonometric identities to simplify load-distribution expressions involving the angle of inclination θ of the ramp. Two expressions appear in her calculations:
E = (sin⁴θ − cos⁴θ) / (sin²θ − cos²θ)
F = (1 − 2cos²θ) / (sin²θ − cos²θ)

A civil engineer is designing a ramp for a warehouse loading dock. To check the structural calculations, she uses trigonometric identities to simplify expressions involving the angle of inclination θ of the ramp.

She writes down the following expression that appears in her load-distribution formula:

E = (sin⁴θ − cos⁴θ) / (sin²θ − cos²θ)

She also notes that for the support beam calculation, a second expression appears:

F = (1 − 2cos²θ) / (sin²θ − cos²θ)

Based on the above context, answer the following:

(i) Show that E simplifies to 1. (1 mark)
(ii) Show that F simplifies to 1. (1 mark)
(iii) If the engineer also needs to verify the identity:
(sinθ + cosθ)² + (sinθ − cosθ)² = 2
prove this identity using standard trigonometric identities.
OR
Simplify the expression: (1 + tan²θ) · cos²θ and state the value it reduces to, giving full working. (2 marks)

Show answer
Part (i) [1 mark]

LHS = (sin⁴θ − cos⁴θ) / (sin²θ − cos²θ)

⟹ Using the difference of squares identity a² − b² = (a − b)(a + b), with a = sin²θ, b = cos²θ:

sin⁴θ − cos⁴θ = (sin²θ − cos²θ)(sin²θ + cos²θ)

⟹ LHS = [(sin²θ − cos²θ)(sin²θ + cos²θ)] / (sin²θ − cos²θ) [since sin²θ ≠ cos²θ, i.e. denominator ≠ 0]

⟹ LHS = sin²θ + cos²θ

By the Pythagorean Identity, sin²θ + cos²θ = 1

⟹ LHS = 1 = RHS

∴ E = 1. Hence proved.

---

Part (ii) [1 mark]

LHS = (1 − 2cos²θ) / (sin²θ − cos²θ)

⟹ Using the identity sin²θ + cos²θ = 1, we write 1 = sin²θ + cos²θ:

Numerator = sin²θ + cos²θ − 2cos²θ = sin²θ − cos²θ

⟹ LHS = (sin²θ − cos²θ) / (sin²θ − cos²θ) [since denominator ≠ 0]

⟹ LHS = 1

∴ F = 1. Hence proved.

---

Part (iii) [2 marks]

Option A: Prove (sinθ + cosθ)² + (sinθ − cosθ)² = 2

LHS = (sinθ + cosθ)² + (sinθ − cosθ)²

⟹ Expanding using (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b²:

= (sin²θ + 2 sinθ cosθ + cos²θ) + (sin²θ − 2 sinθ cosθ + cos²θ)

⟹ = sin²θ + cos²θ + 2 sinθ cosθ + sin²θ + cos²θ − 2 sinθ cosθ

⟹ = (sin²θ + cos²θ) + (sin²θ + cos²θ)

⟹ Using the Pythagorean Identity sin²θ + cos²θ = 1:

= 1 + 1

⟹ = 2 = RHS

∴ (sinθ + cosθ)² + (sinθ − cosθ)² = 2. Hence proved.

---

Option B (OR): Simplify (1 + tan²θ) · cos²θ

Using the identity 1 + tan²θ = sec²θ (Pythagorean Identity):

(1 + tan²θ) · cos²θ = sec²θ · cos²θ

⟹ Since sec θ = 1/cos θ, we have sec²θ = 1/cos²θ:

= (1/cos²θ) · cos²θ

⟹ = 1

∴ (1 + tan²θ) · cos²θ = 1.
Q6Case-based4 marks

A civil engineer is designing a ramp for wheelchair access at a public building. She models the ramp as a right triangle where the angle of inclination is θ, and uses trigonometric identities to verify her structural calculations.

A civil engineer is designing a ramp for a wheelchair access at a public building. To ensure the ramp meets safety standards, she uses trigonometric ratios and identities to verify structural calculations. She models the ramp as a right triangle where the angle of inclination is θ.

Based on this context, answer the following:

(i) If sin θ = 3/5, find the value of cos θ. (1 mark)

(ii) Using the values found, verify that sin²θ + cos²θ = 1. (1 mark)

(iii) The engineer also needs to check the expression: (1 + tan²θ) × cos²θ. Show that this expression always equals 1, for any angle θ, using a trigonometric identity. (2 marks)

Show answer
(i) Given: sin θ = 3/5

Using the identity sin²θ + cos²θ = 1:

⟹ cos²θ = 1 − sin²θ

⟹ cos²θ = 1 − (3/5)²

⟹ cos²θ = 1 − 9/25

⟹ cos²θ = 16/25

∴ cos θ = 4/5 (taking positive value, since θ is an acute angle of the ramp)

(ii) Verification that sin²θ + cos²θ = 1:

LHS = sin²θ + cos²θ

⟹ LHS = (3/5)² + (4/5)²

⟹ LHS = 9/25 + 16/25

⟹ LHS = 25/25

⟹ LHS = 1 = RHS

∴ sin²θ + cos²θ = 1 is verified.

(iii) To show: (1 + tan²θ) × cos²θ = 1

Proof:

LHS = (1 + tan²θ) × cos²θ

Using the identity 1 + tan²θ = sec²θ:

⟹ LHS = sec²θ × cos²θ

Since sec θ = 1/cos θ, we have sec²θ = 1/cos²θ:

⟹ LHS = (1/cos²θ) × cos²θ

⟹ LHS = 1

= RHS

Hence proved.
Q7Case-based4 marks

A civil engineer is designing a ramp for a heritage site. To verify the structural integrity of the ramp, she models the angle of inclination as θ and uses the following trigonometric expression to check load distribution:

E = (sin θ + cosec θ)² + (cos θ + sec θ)²

She claims that the value of E is always at least 9, regardless of the angle θ (where sin θ ≠ 0 and cos θ ≠ 0).

A civil engineer is designing a ramp for a heritage site. To verify the structural integrity of the ramp, she models the angle of inclination as θ and uses the following trigonometric expression to check load distribution:

E = (sin θ + cosec θ)² + (cos θ + sec θ)²

She claims that the value of E is always at least 9, regardless of the angle θ (where sin θ ≠ 0 and cos θ ≠ 0).

(i) Show that E = 7 + tan²θ + cot²θ. [2 marks]
(ii) Using the result of part (i), verify the engineer's claim that E ≥ 9 for all valid values of θ. [2 marks]

Show answer
Part (i): [2 marks]

Expanding the LHS:

E = (sin θ + cosec θ)² + (cos θ + sec θ)²

⟹ E = sin²θ + 2·sin θ·cosec θ + cosec²θ + cos²θ + 2·cos θ·sec θ + sec²θ

Using sin θ · cosec θ = 1 and cos θ · sec θ = 1:

⟹ E = sin²θ + 2(1) + cosec²θ + cos²θ + 2(1) + sec²θ

⟹ E = (sin²θ + cos²θ) + 4 + cosec²θ + sec²θ

Using the identity sin²θ + cos²θ = 1:

⟹ E = 1 + 4 + cosec²θ + sec²θ

Using the identities cosec²θ = 1 + cot²θ and sec²θ = 1 + tan²θ:

⟹ E = 5 + (1 + cot²θ) + (1 + tan²θ)

⟹ E = 7 + tan²θ + cot²θ

= RHS. Hence proved.

---

Part (ii): [2 marks]

From part (i), E = 7 + tan²θ + cot²θ.

Using the AM–GM inequality (or the identity (a − b)² ≥ 0):

For any real number a, a² ≥ 0.

⟹ (tan θ − cot θ)² ≥ 0

⟹ tan²θ − 2·tan θ·cot θ + cot²θ ≥ 0

Using tan θ · cot θ = 1:

⟹ tan²θ + cot²θ − 2 ≥ 0

⟹ tan²θ + cot²θ ≥ 2

Substituting into the expression for E:

⟹ E = 7 + tan²θ + cot²θ ≥ 7 + 2

∴ E ≥ 9

Hence, the engineer's claim is verified — the value of E is always at least 9 for all valid values of θ.
Q8Case-based4 marks

A civil engineer is designing a ramp for a flyover. The structural analysis requires verifying trigonometric identities of the form (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ, where θ is the angle of inclination of the ramp. She needs to analyse properties of this expression for different values of θ.

A civil engineer is designing a ramp for a flyover. The structural analysis requires verifying certain trigonometric ratios. During calculations, she uses the identity:

(sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ

(i) If tan θ + cot θ = k, express tan²θ + cot²θ in terms of k. [1 mark]
(ii) Using the identity above, find the minimum value of (sin θ + cosec θ)² + (cos θ + sec θ)². [1 mark]
(iii) Prove the identity: (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ. [2 marks]

OR

(iii) If sin θ + cos θ = √2 cos θ, prove that cos θ − sin θ = √2 sin θ. [2 marks]

Show answer
(i) Given: tan θ + cot θ = k

Squaring both sides:
⟹ (tan θ + cot θ)² = k²
⟹ tan²θ + 2·tan θ·cot θ + cot²θ = k²
⟹ tan²θ + 2(1) + cot²θ = k² [∵ tan θ · cot θ = 1]
⟹ tan²θ + cot²θ = k² − 2

∴ tan²θ + cot²θ = k² − 2

---

(ii) From the given identity:
(sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ

Now, tan²θ ≥ 0 and cot²θ ≥ 0 for all valid θ.

Also, by AM–GM inequality: tan²θ + cot²θ ≥ 2√(tan²θ · cot²θ) = 2

∴ Minimum value of tan²θ + cot²θ = 2 (attained when tan θ = cot θ, i.e., θ = 45°)

⟹ Minimum value of (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + 2 = 9

∴ The minimum value is 9.

---

(iii) PROOF:

LHS = (sin θ + cosec θ)² + (cos θ + sec θ)²

⟹ = sin²θ + 2·sin θ·cosec θ + cosec²θ + cos²θ + 2·cos θ·sec θ + sec²θ

⟹ = sin²θ + 2(sin θ · 1/sin θ) + cosec²θ + cos²θ + 2(cos θ · 1/cos θ) + sec²θ

⟹ = sin²θ + 2 + cosec²θ + cos²θ + 2 + sec²θ

⟹ = (sin²θ + cos²θ) + 4 + cosec²θ + sec²θ

⟹ = 1 + 4 + cosec²θ + sec²θ [∵ sin²θ + cos²θ = 1]

⟹ = 5 + (1 + cot²θ) + (1 + tan²θ) [∵ cosec²θ = 1 + cot²θ and sec²θ = 1 + tan²θ]

⟹ = 5 + 1 + cot²θ + 1 + tan²θ

⟹ = 7 + tan²θ + cot²θ

= RHS. Hence proved.

---

OR

(iii) PROOF:

Given: sin θ + cos θ = √2 cos θ
To Prove: cos θ − sin θ = √2 sin θ

Starting from the given condition:
sin θ + cos θ = √2 cos θ

⟹ sin θ = √2 cos θ − cos θ

⟹ sin θ = cos θ(√2 − 1)

⟹ cos θ = sin θ / (√2 − 1)

Rationalising the denominator:

⟹ cos θ = sin θ(√2 + 1) / [(√2 − 1)(√2 + 1)]

⟹ cos θ = sin θ(√2 + 1) / (2 − 1)

⟹ cos θ = sin θ(√2 + 1)

Now consider cos θ − sin θ:

⟹ cos θ − sin θ = sin θ(√2 + 1) − sin θ

⟹ = sin θ(√2 + 1 − 1)

⟹ = sin θ · √2

⟹ = √2 sin θ

= RHS. Hence proved.
Q9MCQ1 mark

If sin θ = 3/5, find the value of cos²θ + sin²θ.

Show answer
Option (D) is correct.

Explanation: By the fundamental trigonometric identity, sin²θ + cos²θ = 1 for all values of θ. This identity holds true for any angle θ regardless of the value of sin θ. Therefore, cos²θ + sin²θ = 1.
Q10MCQ1 mark

If sin θ = 3/5, find the value of cos²θ + sin²θ.

Show answer
Option (C) is correct. Explanation: By the fundamental trigonometric identity sin²θ + cos²θ = 1, the expression cos²θ + sin²θ = 1 for all values of θ, regardless of the value of sin θ. The given value sin θ = 3/5 is not needed, as this identity holds universally. ∴ cos²θ + sin²θ = 1.
Q11MCQ1 mark

If sin θ = 3/5, find the value of cos²θ + sin²θ.

Show answer
Option (D) is correct.

Explanation: By the fundamental trigonometric identity sin²θ + cos²θ = 1, the sum cos²θ + sin²θ = 1 for all values of θ, regardless of the value of sin θ. Hence, even though sin θ = 3/5 is given, the identity holds universally and the required value is 1.
Q12MCQ1 mark

If sin θ = 3/5, find the value of cos²θ + sin²θ.

Show answer
Option (C) is correct.

Explanation: By the fundamental trigonometric identity sin²θ + cos²θ = 1, the sum cos²θ + sin²θ equals 1 for all values of θ, regardless of the value of sin θ. The given value sin θ = 3/5 is not required to evaluate this expression, as the identity holds universally. ∴ cos²θ + sin²θ = 1.
Q13MCQ1 mark

If sin θ = 3/5, find the value of cos²θ + sin²θ.

Show answer
Option (C) is correct.

Explanation: By the fundamental trigonometric identity, sin²θ + cos²θ = 1 for all values of θ. This identity holds true regardless of the value of sin θ. Therefore, cos²θ + sin²θ = 1.
Q14MCQ1 mark

If sin θ = 3/5, find the value of cos²θ + sin²θ.

Show answer
Option (C) is correct.

Explanation: By the fundamental trigonometric identity, sin²θ + cos²θ = 1 for all values of θ. This identity holds regardless of the value of sin θ, so even though sin θ = 3/5 is given, the expression cos²θ + sin²θ = 1 directly by identity. No substitution or further calculation is needed.
Q15Short Answer3 marks

Prove that: (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ

Show answer
Prove that: (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ

Taking the Left Hand Side (LHS):

LHS = (sin θ + cosec θ)² + (cos θ + sec θ)²

⟹ = sin²θ + 2·sin θ·cosec θ + cosec²θ + cos²θ + 2·cos θ·sec θ + sec²θ

[Using (a + b)² = a² + 2ab + b²]

⟹ = sin²θ + 2·sin θ·(1/sin θ) + cosec²θ + cos²θ + 2·cos θ·(1/cos θ) + sec²θ

[∵ cosec θ = 1/sin θ and sec θ = 1/cos θ]

⟹ = sin²θ + 2 + cosec²θ + cos²θ + 2 + sec²θ

⟹ = (sin²θ + cos²θ) + cosec²θ + sec²θ + 4

⟹ = 1 + cosec²θ + sec²θ + 4

[∵ sin²θ + cos²θ = 1]

⟹ = 5 + (1 + cot²θ) + (1 + tan²θ)

[∵ cosec²θ = 1 + cot²θ and sec²θ = 1 + tan²θ]

⟹ = 5 + 1 + cot²θ + 1 + tan²θ

⟹ = 7 + tan²θ + cot²θ

= RHS. Hence proved.
Q16Short Answer3 marks

Prove that: (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ

Show answer
Proof:

Taking the LHS:

LHS = (sin θ + cosec θ)² + (cos θ + sec θ)²

⟹ = sin²θ + 2·sin θ·cosec θ + cosec²θ + cos²θ + 2·cos θ·sec θ + sec²θ

⟹ = sin²θ + 2·sin θ·(1/sin θ) + cosec²θ + cos²θ + 2·cos θ·(1/cos θ) + sec²θ

⟹ = sin²θ + 2 + cosec²θ + cos²θ + 2 + sec²θ

⟹ = (sin²θ + cos²θ) + cosec²θ + sec²θ + 4

Using the identity sin²θ + cos²θ = 1:

⟹ = 1 + cosec²θ + sec²θ + 4

⟹ = 5 + cosec²θ + sec²θ

Now applying the identities cosec²θ = 1 + cot²θ and sec²θ = 1 + tan²θ:

⟹ = 5 + (1 + cot²θ) + (1 + tan²θ)

⟹ = 5 + 2 + tan²θ + cot²θ

⟹ = 7 + tan²θ + cot²θ

= RHS. Hence proved.
Q17Short Answer3 marks

Prove that: (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ

Show answer
Proof:

Taking the LHS:

LHS = (sin θ + cosec θ)² + (cos θ + sec θ)²

⟹ = sin²θ + 2·sin θ·cosec θ + cosec²θ + cos²θ + 2·cos θ·sec θ + sec²θ

⟹ = sin²θ + 2·sin θ·(1/sin θ) + cosec²θ + cos²θ + 2·cos θ·(1/cos θ) + sec²θ

[∵ cosec θ = 1/sin θ and sec θ = 1/cos θ]

⟹ = sin²θ + 2 + cosec²θ + cos²θ + 2 + sec²θ

⟹ = (sin²θ + cos²θ) + 4 + cosec²θ + sec²θ

Using the identity sin²θ + cos²θ = 1:

⟹ = 1 + 4 + cosec²θ + sec²θ

Using the identities cosec²θ = 1 + cot²θ and sec²θ = 1 + tan²θ:

⟹ = 5 + (1 + cot²θ) + (1 + tan²θ)

⟹ = 5 + 1 + cot²θ + 1 + tan²θ

⟹ = 7 + tan²θ + cot²θ

= RHS. Hence proved.
Q18Short Answer3 marks

Prove that: (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ

Show answer
Proof:

Taking the Left Hand Side (LHS):

LHS = (sin θ + cosec θ)² + (cos θ + sec θ)²

⟹ = sin²θ + 2·sin θ·cosec θ + cosec²θ + cos²θ + 2·cos θ·sec θ + sec²θ

⟹ = sin²θ + cos²θ + 2·sin θ·(1/sin θ) + 2·cos θ·(1/cos θ) + cosec²θ + sec²θ

⟹ = sin²θ + cos²θ + 2(1) + 2(1) + cosec²θ + sec²θ

Using the identity sin²θ + cos²θ = 1:

⟹ = 1 + 2 + 2 + cosec²θ + sec²θ

⟹ = 5 + cosec²θ + sec²θ

Now, using the identities cosec²θ = 1 + cot²θ and sec²θ = 1 + tan²θ:

⟹ = 5 + (1 + cot²θ) + (1 + tan²θ)

⟹ = 5 + 1 + 1 + tan²θ + cot²θ

⟹ = 7 + tan²θ + cot²θ

= RHS. Hence proved.
Q19Short Answer3 marks

Prove that: (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ

Show answer
Proof:

Taking the L.H.S.:

L.H.S. = (sin θ + cosec θ)² + (cos θ + sec θ)²

⟹ = sin²θ + 2·sin θ·cosec θ + cosec²θ + cos²θ + 2·cos θ·sec θ + sec²θ

[Using (a + b)² = a² + 2ab + b²]

⟹ = sin²θ + cosec²θ + 2·sin θ·(1/sin θ) + cos²θ + sec²θ + 2·cos θ·(1/cos θ)

[∵ cosec θ = 1/sin θ and sec θ = 1/cos θ]

⟹ = sin²θ + cosec²θ + 2 + cos²θ + sec²θ + 2

⟹ = (sin²θ + cos²θ) + cosec²θ + sec²θ + 4

⟹ = 1 + cosec²θ + sec²θ + 4

[∵ sin²θ + cos²θ = 1]

⟹ = 5 + (1 + cot²θ) + (1 + tan²θ)

[∵ cosec²θ = 1 + cot²θ and sec²θ = 1 + tan²θ]

⟹ = 5 + 1 + cot²θ + 1 + tan²θ

⟹ = 7 + tan²θ + cot²θ

= R.H.S.

Hence proved.
Q20Short Answer3 marks

Prove that: (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ

Show answer
Proof:

Taking the LHS:

LHS = (sin θ + cosec θ)² + (cos θ + sec θ)²

⟹ = sin²θ + 2·sin θ·cosec θ + cosec²θ + cos²θ + 2·cos θ·sec θ + sec²θ

⟹ = sin²θ + cos²θ + 2·sin θ·(1/sin θ) + cosec²θ + 2·cos θ·(1/cos θ) + sec²θ

⟹ = (sin²θ + cos²θ) + 2(1) + cosec²θ + 2(1) + sec²θ

Using the identity sin²θ + cos²θ = 1:

⟹ = 1 + 2 + 2 + cosec²θ + sec²θ

⟹ = 5 + cosec²θ + sec²θ

Using the identities sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ:

⟹ = 5 + (1 + cot²θ) + (1 + tan²θ)

⟹ = 5 + 1 + 1 + tan²θ + cot²θ

⟹ = 7 + tan²θ + cot²θ

= RHS. Hence proved.
Q21Case-based4 marks

A civil engineer is designing a ramp for wheelchair access at a community centre. She models the ramp as a right triangle. For the angle of inclination θ, she records sin θ = 5/13.

A civil engineer is designing a ramp for wheelchair access at a community centre. To check the structural integrity, she models the ramp as a right triangle and uses trigonometric ratios. For the angle of inclination θ of the ramp, she records:

sin θ = 5/13

Using this information, answer the following questions:

(i) Find the value of cos θ. [1 mark]
(ii) Find the value of tan θ. [1 mark]
(iii) Prove that: (1 − sin²θ) × sec²θ = 1 [2 marks]

OR

(iii) Prove that: (sin θ + cos θ)² + (sin θ − cos θ)² = 2 [2 marks]

Show answer
(i) Finding cos θ:

Using the identity sin²θ + cos²θ = 1:

⟹ cos²θ = 1 − sin²θ

⟹ cos²θ = 1 − (5/13)²

⟹ cos²θ = 1 − 25/169

⟹ cos²θ = 144/169

∴ cos θ = 12/13

(Since θ is an acute angle, cos θ > 0.)

(ii) Finding tan θ:

tan θ = sin θ / cos θ

⟹ tan θ = (5/13) ÷ (12/13)

⟹ tan θ = (5/13) × (13/12)

∴ tan θ = 5/12

(iii) Proof of (1 − sin²θ) × sec²θ = 1:

LHS = (1 − sin²θ) × sec²θ

Using the identity 1 − sin²θ = cos²θ:

⟹ LHS = cos²θ × sec²θ

Using the identity sec θ = 1/cos θ, so sec²θ = 1/cos²θ:

⟹ LHS = cos²θ × (1/cos²θ)

⟹ LHS = 1

= RHS. Hence proved.

OR

(iii) Proof of (sin θ + cos θ)² + (sin θ − cos θ)² = 2:

LHS = (sin θ + cos θ)² + (sin θ − cos θ)²

Expanding using (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b²:

⟹ LHS = (sin²θ + 2 sin θ cos θ + cos²θ) + (sin²θ − 2 sin θ cos θ + cos²θ)

⟹ LHS = sin²θ + cos²θ + 2 sin θ cos θ + sin²θ + cos²θ − 2 sin θ cos θ

⟹ LHS = (sin²θ + cos²θ) + (sin²θ + cos²θ)

Using the identity sin²θ + cos²θ = 1:

⟹ LHS = 1 + 1

⟹ LHS = 2

= RHS. Hence proved.
Q22Case-based4 marks

A civil engineer is designing a ramp for a building entrance. To verify the structural calculations, she needs to confirm that a particular trigonometric expression simplifies correctly. She writes the following expressions on her notepad and needs to verify each result step by step.

A civil engineer is designing a ramp for a building entrance. To verify the structural calculations, she needs to confirm that a particular trigonometric expression simplifies correctly. She writes the following expression on her notepad:

(sin θ + cos θ)² + (sin θ − cos θ)²

She also separately notes the expression: (1 + tan²θ) · cos²θ

(i) Evaluate (sin θ + cos θ)² + (sin θ − cos θ)². [1]
(ii) Evaluate (1 + tan²θ) · cos²θ. [1]
(iii) The engineer now needs to prove the following identity for her report:

(sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ

Prove the above identity. [2]

OR

(iii) Prove that: (tan θ + sin θ)/(tan θ − sin θ) = (sec θ + 1)/(sec θ − 1). [2]

Show answer
(i) Evaluate (sin θ + cos θ)² + (sin θ − cos θ)²

Expanding using (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b²:

⟹ (sin²θ + 2 sin θ cos θ + cos²θ) + (sin²θ − 2 sin θ cos θ + cos²θ)

⟹ (sin²θ + cos²θ) + (sin²θ + cos²θ)

Using the identity sin²θ + cos²θ = 1:

⟹ 1 + 1

∴ (sin θ + cos θ)² + (sin θ − cos θ)² = 2

(ii) Evaluate (1 + tan²θ) · cos²θ

Using the identity 1 + tan²θ = sec²θ:

⟹ sec²θ · cos²θ

⟹ (1/cos²θ) · cos²θ

∴ (1 + tan²θ) · cos²θ = 1

(iii) Prove: (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ

Taking LHS:

LHS = (sin θ + cosec θ)² + (cos θ + sec θ)²

⟹ sin²θ + 2 sin θ · cosec θ + cosec²θ + cos²θ + 2 cos θ · sec θ + sec²θ

⟹ sin²θ + 2 sin θ · (1/sin θ) + cosec²θ + cos²θ + 2 cos θ · (1/cos θ) + sec²θ

⟹ sin²θ + 2 + cosec²θ + cos²θ + 2 + sec²θ

⟹ (sin²θ + cos²θ) + 4 + cosec²θ + sec²θ

Using sin²θ + cos²θ = 1:

⟹ 1 + 4 + cosec²θ + sec²θ

Using identities sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ:

⟹ 5 + (1 + cot²θ) + (1 + tan²θ)

⟹ 5 + 1 + 1 + tan²θ + cot²θ

⟹ 7 + tan²θ + cot²θ

= RHS. Hence proved.

OR

(iii) Prove: (tan θ + sin θ)/(tan θ − sin θ) = (sec θ + 1)/(sec θ − 1)

Taking LHS:

LHS = (tan θ + sin θ)/(tan θ − sin θ)

⟹ (sin θ/cos θ + sin θ)/(sin θ/cos θ − sin θ)

⟹ [sin θ(1/cos θ + 1)] / [sin θ(1/cos θ − 1)]

Cancelling sin θ (sin θ ≠ 0) from numerator and denominator:

⟹ (1/cos θ + 1)/(1/cos θ − 1)

⟹ (sec θ + 1)/(sec θ − 1)

= RHS. Hence proved.
Q23Case-based4 marks

A civil engineer is designing a ramp for a building entrance. To verify the structural calculations, she uses trigonometric identities involving the angle of inclination θ of the ramp. She needs to confirm several algebraic-trigonometric expressions to ensure the design equations are consistent.

A civil engineer is designing a ramp for a building entrance. To verify the structural calculations, she uses the following trigonometric expression involving the angle of inclination θ of the ramp:

She needs to confirm that:

(sin θ + cos θ)² + (sin θ − cos θ)² = 2

(i) Prove the above identity. [1 mark]

(ii) The engineer also notes that for the ramp angle θ, it is given that tan θ + cot θ = 2. Using this, find the value of tan²θ + cot²θ. [1 mark]

(iii) For a different section of the ramp, the engineer needs to simplify the expression:

(sin⁴θ − cos⁴θ) / (sin²θ − cos²θ)

Prove that this expression equals 1, for all valid values of θ. [2 marks]

OR

(iii) Prove that: (1 + cot θ − cosec θ)(1 + tan θ + sec θ) = 2. [2 marks]

Show answer
(i) LHS = (sin θ + cos θ)² + (sin θ − cos θ)²
⟹ = (sin²θ + 2 sin θ cos θ + cos²θ) + (sin²θ − 2 sin θ cos θ + cos²θ)
⟹ = (sin²θ + cos²θ) + 2 sin θ cos θ + (sin²θ + cos²θ) − 2 sin θ cos θ
⟹ = 1 + 2 sin θ cos θ + 1 − 2 sin θ cos θ [Using sin²θ + cos²θ = 1]
⟹ = 2
= RHS. Hence proved.

(ii) Given: tan θ + cot θ = 2
Squaring both sides:
⟹ (tan θ + cot θ)² = 4
⟹ tan²θ + 2 tan θ · cot θ + cot²θ = 4
⟹ tan²θ + 2(1) + cot²θ = 4 [∵ tan θ · cot θ = 1]
⟹ tan²θ + cot²θ = 4 − 2
∴ tan²θ + cot²θ = 2

(iii) [Main Option]
LHS = (sin⁴θ − cos⁴θ) / (sin²θ − cos²θ)

Using the identity a² − b² = (a + b)(a − b), with a = sin²θ and b = cos²θ:
⟹ sin⁴θ − cos⁴θ = (sin²θ − cos²θ)(sin²θ + cos²θ)

∴ LHS = [(sin²θ − cos²θ)(sin²θ + cos²θ)] / (sin²θ − cos²θ)
⟹ = sin²θ + cos²θ [∵ sin²θ − cos²θ ≠ 0 for valid θ]
⟹ = 1 [Using the identity sin²θ + cos²θ = 1]
= RHS. Hence proved.

[OR]

(iii) [Alternate Option]
LHS = (1 + cot θ − cosec θ)(1 + tan θ + sec θ)

Express everything in terms of sin θ and cos θ:
⟹ = (1 + cos θ/sin θ − 1/sin θ)(1 + sin θ/cos θ + 1/cos θ)
⟹ = [(sin θ + cos θ − 1)/sin θ] × [(cos θ + sin θ + 1)/cos θ]
⟹ = [(sin θ + cos θ − 1)(sin θ + cos θ + 1)] / (sin θ cos θ)

Let p = sin θ + cos θ. Numerator = (p − 1)(p + 1) = p² − 1:
⟹ = [(sin θ + cos θ)² − 1] / (sin θ cos θ)
⟹ = [sin²θ + 2 sin θ cos θ + cos²θ − 1] / (sin θ cos θ)
⟹ = [1 + 2 sin θ cos θ − 1] / (sin θ cos θ) [∵ sin²θ + cos²θ = 1]
⟹ = [2 sin θ cos θ] / (sin θ cos θ)
⟹ = 2
= RHS. Hence proved.
Q24Case-based4 marks

A civil engineer is designing a ramp for a building entrance. While checking structural calculations, she notices the expression (sin θ + cos θ)² + (sin θ − cos θ)² appearing repeatedly in her trigonometric model. She claims this expression always equals 2, regardless of the angle θ, used.

A civil engineer is designing a ramp for a building entrance. While checking the structural calculations, she notices the following expression appearing repeatedly in her trigonometric model:

(sin θ + cos θ)² + (sin θ − cos θ)²

She claims this expression always equals 2, regardless of the angle θ, which simplifies her calculations greatly.

Based on this context, answer the following:

(i) Expand (sin θ + cos θ)² and write the result. [1]
(ii) Expand (sin θ − cos θ)² and write the result. [1]
(iii) Using parts (i) and (ii), prove that (sin θ + cos θ)² + (sin θ − cos θ)² = 2, and hence verify the engineer's claim by checking the result for θ = 45°.

OR

(iii) A student simplifies the expression (1 + tan²θ) × cos²θ. Using a suitable trigonometric identity, simplify it fully and find its value when θ = 60°. [2]

Show answer
(i) Expanding (sin θ + cos θ)²:

(sin θ + cos θ)² = sin²θ + 2 sin θ cos θ + cos²θ

∴ (sin θ + cos θ)² = 1 + 2 sin θ cos θ

[Using the identity sin²θ + cos²θ = 1]

(ii) Expanding (sin θ − cos θ)²:

(sin θ − cos θ)² = sin²θ − 2 sin θ cos θ + cos²θ

∴ (sin θ − cos θ)² = 1 − 2 sin θ cos θ

[Using the identity sin²θ + cos²θ = 1]

(iii) Adding the results from (i) and (ii):

LHS = (sin θ + cos θ)² + (sin θ − cos θ)²

⟹ LHS = (1 + 2 sin θ cos θ) + (1 − 2 sin θ cos θ)

⟹ LHS = 1 + 1 + 2 sin θ cos θ − 2 sin θ cos θ

⟹ LHS = 2

= RHS. Hence proved.

Verification for θ = 45°:

Using sin 45° = 1/√2 and cos 45° = 1/√2,

(sin 45° + cos 45°)² + (sin 45° − cos 45°)²

⟹ (1/√2 + 1/√2)² + (1/√2 − 1/√2)²

⟹ (2/√2)² + (0)²

⟹ (√2)² + 0

⟹ 2 + 0 = 2

∴ The expression equals 2 at θ = 45°, confirming the engineer's claim.

OR

(iii) The expression to simplify is (1 + tan²θ) × cos²θ.

Using the identity 1 + tan²θ = sec²θ,

(1 + tan²θ) × cos²θ = sec²θ × cos²θ

⟹ (1/cos²θ) × cos²θ

⟹ 1

∴ (1 + tan²θ) × cos²θ = 1 for all values of θ.

For θ = 60°: Since the expression simplifies to the constant 1,

∴ (1 + tan²60°) × cos²60° = 1.
Q25Case-based4 marks

A civil engineer is designing a ramp for a flyover. She models the cross-section of the ramp as a right triangle where the angle of inclination is θ. To verify structural stability, she needs to confirm certain trigonometric relationships. She writes:
P = (sin θ + cos θ)² + (sin θ − cos θ)²
Q = (sec²θ − 1)(1 − sin²θ)
She also notes that for a specific inclination, tan θ + cot θ = 2.

A civil engineer is designing a ramp for a flyover. She models the cross-section of the ramp as a right triangle where the angle of inclination is θ. To verify the structural stability, she needs to confirm certain trigonometric relationships that hold for any angle θ. While checking her calculations, she writes the following expressions:

P = (sin θ + cos θ)² + (sin θ − cos θ)²
Q = (sec²θ − 1)(1 − sin²θ)

She also notes that for a specific inclination, tan θ + cot θ = 2.

Based on the above information, answer the following:

(i) Find the value of P. [1 mark]
(ii) Find the value of Q. [1 mark]
(iii) If tan θ + cot θ = 2, find the value of tan²θ + cot²θ. [2 marks]
OR
Prove the identity: (1 + tan²θ) / (1 + cot²θ) = tan²θ

Show answer
Part (i): Find the value of P.

P = (sin θ + cos θ)² + (sin θ − cos θ)²

Using the identity (a + b)² + (a − b)² = 2(a² + b²):

⟹ P = 2(sin²θ + cos²θ)

Using the identity sin²θ + cos²θ = 1:

⟹ P = 2(1)

∴ P = 2

─────────────────────────────────────

Part (ii): Find the value of Q.

Q = (sec²θ − 1)(1 − sin²θ)

Using the identity sec²θ − 1 = tan²θ and 1 − sin²θ = cos²θ:

⟹ Q = tan²θ × cos²θ

⟹ Q = (sin²θ / cos²θ) × cos²θ

⟹ Q = sin²θ

∴ Q = sin²θ

(Note: This simplifies to a value that depends on θ, but in its simplest closed form, Q = sin²θ. The engineer confirms it is always between 0 and 1, ensuring a valid structural parameter.)

─────────────────────────────────────

Part (iii) [Main option]: If tan θ + cot θ = 2, find tan²θ + cot²θ.

We use the identity:

(tan θ + cot θ)² = tan²θ + 2·tan θ·cot θ + cot²θ

⟹ (tan θ + cot θ)² = tan²θ + cot²θ + 2·(tan θ · cot θ)

Now, tan θ · cot θ = tan θ × (1/tan θ) = 1

⟹ (2)² = tan²θ + cot²θ + 2(1)

⟹ 4 = tan²θ + cot²θ + 2

⟹ tan²θ + cot²θ = 4 − 2

∴ tan²θ + cot²θ = 2

─────────────────────────────────────

Part (iii) [OR option]: Prove that (1 + tan²θ) / (1 + cot²θ) = tan²θ

LHS = (1 + tan²θ) / (1 + cot²θ)

Using identities 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ:

⟹ LHS = sec²θ / cosec²θ

⟹ LHS = (1/cos²θ) / (1/sin²θ)

⟹ LHS = (1/cos²θ) × (sin²θ/1)

⟹ LHS = sin²θ / cos²θ

⟹ LHS = tan²θ

= RHS. Hence proved.
Q26Case-based4 marks

A structural engineer is designing a support bracket. While verifying the stress equations, she encounters trigonometric expressions that must simplify to specific constants for the design to be structurally safe. She uses fundamental trigonometric identities to verify each expression.

A structural engineer is designing a support bracket. While verifying the stress equations, she encounters the following trigonometric expression that must simplify to a constant for the design to be safe:

(sin³θ + cos³θ)/(sinθ + cosθ) + sinθ cosθ

(i) Prove that the expression above is equal to 1. [2 marks]
(ii) Using the result from part (i), or otherwise, prove that:
(tanθ + cotθ)(sinθ + cosθ) = (secθ + cosecθ) [2 marks]

Show answer
Part (i): Prove that (sin³θ + cos³θ)/(sinθ + cosθ) + sinθ cosθ = 1

Taking LHS:

= (sin³θ + cos³θ)/(sinθ + cosθ) + sinθ cosθ

Using the algebraic identity a³ + b³ = (a + b)(a² − ab + b²), with a = sinθ, b = cosθ:

⟹ sin³θ + cos³θ = (sinθ + cosθ)(sin²θ − sinθ cosθ + cos²θ)

Substituting:

⟹ LHS = [(sinθ + cosθ)(sin²θ − sinθ cosθ + cos²θ)]/(sinθ + cosθ) + sinθ cosθ

⟹ = (sin²θ − sinθ cosθ + cos²θ) + sinθ cosθ

Using the identity sin²θ + cos²θ = 1:

⟹ = 1 − sinθ cosθ + sinθ cosθ

⟹ = 1

= RHS. Hence proved.

---

Part (ii): Prove that (tanθ + cotθ)(sinθ + cosθ) = (secθ + cosecθ)

Taking LHS:

= (tanθ + cotθ)(sinθ + cosθ)

Expressing tanθ and cotθ in terms of sinθ and cosθ:

⟹ = (sinθ/cosθ + cosθ/sinθ)(sinθ + cosθ)

⟹ = [(sin²θ + cos²θ)/(sinθ cosθ)](sinθ + cosθ)

Using sin²θ + cos²θ = 1:

⟹ = (sinθ + cosθ)/(sinθ cosθ)

⟹ = sinθ/(sinθ cosθ) + cosθ/(sinθ cosθ)

⟹ = 1/cosθ + 1/sinθ

⟹ = secθ + cosecθ

= RHS. Hence proved.
Q27Case-based4 marks

A civil engineer is designing a ramp for a heritage site. She models the ramp surface using two trigonometric expressions E and F (defined in the question) and claims both simplify to the same constant value.

A civil engineer is designing a ramp for a heritage site. She models the ramp surface using the expression:

E = (sin θ + cos θ)² + (sin θ − cos θ)²

and separately computes:

F = (1 + cot θ − cosec θ)(1 + tan θ + sec θ)

She claims that both E and F simplify to the same constant value, independent of θ.

(i) Simplify E and find its value. [1 mark]
(ii) Simplify F and find its value. [2 marks]
(iii) Hence verify her claim, and also prove that:

(tan θ + sin θ)/(tan θ − sin θ) = (sec θ + 1)/(sec θ − 1)

[1 mark]

Show answer
(i) Simplify E:

E = (sin θ + cos θ)² + (sin θ − cos θ)²

Using the identities (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b²:

⟹ E = (sin²θ + 2 sin θ cos θ + cos²θ) + (sin²θ − 2 sin θ cos θ + cos²θ)

⟹ E = (sin²θ + cos²θ) + 2 sin θ cos θ + (sin²θ + cos²θ) − 2 sin θ cos θ

Using the identity sin²θ + cos²θ = 1:

⟹ E = 1 + 2 sin θ cos θ + 1 − 2 sin θ cos θ

∴ E = 2

(ii) Simplify F:

F = (1 + cot θ − cosec θ)(1 + tan θ + sec θ)

Express all ratios in terms of sin θ and cos θ:

⟹ F = (1 + cos θ/sin θ − 1/sin θ)(1 + sin θ/cos θ + 1/cos θ)

⟹ F = [(sin θ + cos θ − 1)/sin θ] × [(cos θ + sin θ + 1)/cos θ]

⟹ F = [(sin θ + cos θ − 1)(sin θ + cos θ + 1)] / (sin θ · cos θ)

Let P = sin θ + cos θ, so the numerator becomes (P − 1)(P + 1) = P² − 1:

⟹ Numerator = (sin θ + cos θ)² − 1

⟹ = sin²θ + 2 sin θ cos θ + cos²θ − 1

⟹ = 1 + 2 sin θ cos θ − 1 [using sin²θ + cos²θ = 1]

⟹ = 2 sin θ cos θ

∴ F = 2 sin θ cos θ / (sin θ · cos θ) = 2

∴ F = 2

(iii) Since E = 2 and F = 2, both expressions simplify to the same constant value 2, verifying the engineer's claim.

Now prove: (tan θ + sin θ)/(tan θ − sin θ) = (sec θ + 1)/(sec θ − 1)

Taking LHS:

LHS = (tan θ + sin θ)/(tan θ − sin θ)

Express tan θ = sin θ/cos θ:

⟹ = (sin θ/cos θ + sin θ)/(sin θ/cos θ − sin θ)

⟹ = [sin θ(1/cos θ + 1)] / [sin θ(1/cos θ − 1)]

⟹ = (1/cos θ + 1)/(1/cos θ − 1)

⟹ = (sec θ + 1)/(sec θ − 1)

= RHS. Hence proved.
Q28Case-based4 marks

A civil engineer is designing a triangular support brace for a bridge. She models the stress ratio at a critical joint using the trigonometric expression E = (sin θ + cos θ)² + (sin θ − cos θ)². She claims that E is always constant, regardless of the angle θ of the brace.

A civil engineer is designing a triangular support brace for a bridge. She models the stress ratio at a critical joint using the expression:

E = (sin θ + cos θ)² + (sin θ − cos θ)²

She claims that no matter what angle θ is used in the design, the stress ratio E always equals 2 — a constant — and that this is independent of whether the brace is tilted at 30°, 45°, or 60°.

Based on this context, answer the following:

(i) Prove algebraically that E = (sin θ + cos θ)² + (sin θ − cos θ)² = 2 for all values of θ. (1 mark)

(ii) The engineer now checks a second stress expression at another joint:
F = (1 + tan²θ) · cos²θ
Simplify F and state its value. (1 mark)

(iii) The engineer finds that at a third joint, the load distribution satisfies:
tan θ + cot θ = (sec θ · cosec θ)
Prove that the left-hand side equals the right-hand side, establishing this identity. (2 marks)

Show answer
(i) Prove that E = (sin θ + cos θ)² + (sin θ − cos θ)² = 2

Expanding using the identity (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b²:

E = (sin²θ + 2 sin θ cos θ + cos²θ) + (sin²θ − 2 sin θ cos θ + cos²θ)

⟹ E = (sin²θ + cos²θ) + 2 sin θ cos θ + (sin²θ + cos²θ) − 2 sin θ cos θ

Using the identity sin²θ + cos²θ = 1:

⟹ E = 1 + 2 sin θ cos θ + 1 − 2 sin θ cos θ

⟹ E = 2

∴ E = 2 for all values of θ. Hence proved.

---

(ii) Simplify F = (1 + tan²θ) · cos²θ

Using the identity 1 + tan²θ = sec²θ:

F = sec²θ · cos²θ

⟹ F = (1/cos²θ) · cos²θ

⟹ F = 1

∴ F = 1

---

(iii) Prove that tan θ + cot θ = sec θ · cosec θ

Working on the LHS only:

LHS = tan θ + cot θ

⟹ = (sin θ / cos θ) + (cos θ / sin θ)

⟹ = (sin²θ + cos²θ) / (sin θ · cos θ)

Using the identity sin²θ + cos²θ = 1:

⟹ = 1 / (sin θ · cos θ)

⟹ = (1 / cos θ) · (1 / sin θ)

⟹ = sec θ · cosec θ

= RHS. Hence proved.
Q29Case-based4 marks

A civil engineer is designing a ramp for a heritage bridge restoration project. The ramp makes an angle θ with the horizontal ground. During quality testing, the engineer records that sin θ + cos θ = √2 · sin θ, and uses trigonometric identities to verify structural calculations.

A civil engineer is designing a ramp for a heritage bridge restoration project. The ramp makes an angle θ with the horizontal ground. During quality testing, the engineer records the following trigonometric measurements at angle θ:

• sin θ + cos θ = √2 · sin θ

Using this relationship, the engineer needs to verify several structural calculations involving trigonometric identities.

Based on the above information, answer the following questions:

(i) Show that sin θ − cos θ = √2 · cos θ. [1]

(ii) Using the results from (i) and the given condition, find the value of tan θ. [1]

(iii) The engineer also needs to verify the following identity for structural load calculations:

(sin θ + cos θ)² + (sin θ − cos θ)² = 2

Prove the above identity. [2]

OR

(iii) The engineer uses the formula:

(1 + tan²θ) · cos²θ = 1

Prove this identity. [2]

Show answer
(i) Given: sin θ + cos θ = √2 · sin θ

⟹ cos θ = √2 · sin θ − sin θ

⟹ cos θ = sin θ(√2 − 1)

Multiply both sides by (√2 + 1):

⟹ cos θ · (√2 + 1) = sin θ · (√2 − 1)(√2 + 1)

⟹ cos θ · (√2 + 1) = sin θ · (2 − 1)

⟹ cos θ · (√2 + 1) = sin θ

⟹ √2 · cos θ + cos θ = sin θ

⟹ sin θ − cos θ = √2 · cos θ

∴ sin θ − cos θ = √2 · cos θ. Hence proved.

---

(ii) From the given condition:
sin θ + cos θ = √2 · sin θ ...(i)

From part (i):
sin θ − cos θ = √2 · cos θ ...(ii)

Adding (i) and (ii):

(sin θ + cos θ) + (sin θ − cos θ) = √2 · sin θ + √2 · cos θ

⟹ 2 sin θ = √2(sin θ + cos θ)

Dividing both sides by 2 cos θ:

From equation (ii): sin θ − cos θ = √2 · cos θ

⟹ sin θ = cos θ + √2 · cos θ = cos θ(1 + √2)

⟹ sin θ / cos θ = 1 + √2

∴ tan θ = 1 + √2

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(iii) [Main Option]

To Prove: (sin θ + cos θ)² + (sin θ − cos θ)² = 2

Taking LHS:

LHS = (sin θ + cos θ)² + (sin θ − cos θ)²

Using the identity (a + b)² + (a − b)² = 2(a² + b²):

⟹ LHS = 2(sin²θ + cos²θ)

Using the identity sin²θ + cos²θ = 1:

⟹ LHS = 2 × 1

⟹ LHS = 2

= RHS. Hence proved.

---

(iii) [OR Option]

To Prove: (1 + tan²θ) · cos²θ = 1

Taking LHS:

LHS = (1 + tan²θ) · cos²θ

Using the identity 1 + tan²θ = sec²θ:

⟹ LHS = sec²θ · cos²θ

Using sec θ = 1/cos θ:

⟹ LHS = (1/cos²θ) · cos²θ

⟹ LHS = 1

= RHS. Hence proved.
Q30Case-based4 marks

A civil engineer is designing a ramp for a wheelchair access point. While verifying the structural calculations, she needs to confirm a trigonometric relationship involving the angle of inclination θ of the ramp. She writes down the following expression on her notepad:

(sin θ + cos θ)² + (sin θ − cos θ)²

She claims this expression always equals 2, regardless of the angle θ, which means the structural ratio is constant for any ramp angle.

A civil engineer is designing a ramp for a wheelchair access point. While verifying the structural calculations, she needs to confirm a trigonometric relationship involving the angle of inclination θ of the ramp. She writes down the following expression on her notepad:

(sin θ + cos θ)² + (sin θ − cos θ)²

She claims this expression always equals 2, regardless of the angle θ, which means the structural ratio is constant for any ramp angle.

(i) Expand (sin θ + cos θ)² [1 mark]
(ii) Expand (sin θ − cos θ)² [1 mark]
(iii) Using your results from (i) and (ii), prove that (sin θ + cos θ)² + (sin θ − cos θ)² = 2, and hence verify the engineer's claim. [2 marks]

Show answer
(i) Expanding (sin θ + cos θ)²:

Using the identity (a + b)² = a² + 2ab + b²,

(sin θ + cos θ)² = sin²θ + 2 sin θ cos θ + cos²θ

∴ (sin θ + cos θ)² = sin²θ + 2 sin θ cos θ + cos²θ

[1 mark]

(ii) Expanding (sin θ − cos θ)²:

Using the identity (a − b)² = a² − 2ab + b²,

(sin θ − cos θ)² = sin²θ − 2 sin θ cos θ + cos²θ

∴ (sin θ − cos θ)² = sin²θ − 2 sin θ cos θ + cos²θ

[1 mark]

(iii) Adding the results from (i) and (ii):

LHS = (sin θ + cos θ)² + (sin θ − cos θ)²

⟹ LHS = (sin²θ + 2 sin θ cos θ + cos²θ) + (sin²θ − 2 sin θ cos θ + cos²θ)

⟹ LHS = sin²θ + cos²θ + 2 sin θ cos θ + sin²θ + cos²θ − 2 sin θ cos θ

⟹ LHS = (sin²θ + cos²θ) + (sin²θ + cos²θ) + (2 sin θ cos θ − 2 sin θ cos θ)

Using the fundamental trigonometric identity sin²θ + cos²θ = 1,

⟹ LHS = 1 + 1 + 0

⟹ LHS = 2 = RHS

∴ (sin θ + cos θ)² + (sin θ − cos θ)² = 2. Hence proved.

Since the result equals 2 for all values of θ, the engineer's claim is correct — the expression is constant and independent of the ramp angle θ.

[2 marks]

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