HOTS Questions on Class 10 Maths Chapter 8 ‘Introduction to Trigonometry’ with Answers

Higher-Order Thinking Skills (HOTS) question set for Chapter 8 ‘Introduction to Trigonometry’ for Class 10 is designed to move students beyond memorisation and routine exercises, cultivating deep conceptual understanding and adaptable problem-solving skills in trigonometry. Aligned with CBSE and NCERT objectives, these questions emphasise clear reasoning and creative application of ideas such as defining trigonometric ratios for acute angles, using identities and relationships between ratios, solving right-triangle problems, interpreting trigonometric values graphically, and real-world situations involving heights and distances. Each question includes a clear answer and brief reasoning to support classroom discussion and self-study. Download the PDF to access the same HOTS questions in a printable format, making offline learning and exam revision easier.

Solved HOTS Questions on Coordinate Geometry

Question 1: If cosec θ = 3x and cot θ = 3/x, find the value of (x² − 1/x²).

Solution:

Using the identity cosec²θ − cot²θ = 1:

(3x)² − (3/x)² = 1 9x² − 9/x² = 1

Factor out 9:

9(x² − 1/x²) = 1 x² − 1/x² = 1/9

⇒ x² − 1/x² = 1/9

Question 2: Prove (sinθ − cosθ + 1) / (sinθ + cosθ − 1) = 1 / (secθ − tanθ)

Solution:

Take LHS: (sinθ − cosθ + 1) / (sinθ + cosθ − 1)

Divide numerator and denominator by cosθ:

= (tanθ − 1 + secθ) / (tanθ + 1 − secθ)

Rearrange numerator: (secθ + tanθ − 1). 

Replace the 1 using the identity:

1 = sec²θ − tan²θ = (secθ − tanθ)(secθ + tanθ) 

So: secθ + tanθ − 1 = secθ + tanθ − (secθ − tanθ)(secθ + tanθ) 

= (secθ + tanθ)[1 − (secθ − tanθ)]

Denominator: tanθ + 1 − secθ = 1 − (secθ − tanθ). 

Let S = secθ − tanθ:

LHS = (secθ + tanθ)(1 − S) / (1 − S) = secθ + tanθ

But we need 1/(secθ − tanθ). 

Rationalise:

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

= (sec²θ − tan²θ) / (sec θ − tan θ) = 1 / (sec θ − tan θ) = RHS

LHS = RHS. 

Proved. 

Question 3: Prove that (cosecθ − sinθ)(secθ − cosθ) = 1 / (tanθ + cotθ)

Solution: 

Simplify each bracket on LHS:

cosecθ − sinθ = 1/sinθ − sinθ 

= (1 − sin²θ)/sinθ 

= cos²θ/sinθ secθ − cosθ 

= 1/cosθ − cosθ = (1 − cos²θ)/cosθ = sin²θ/cosθ

Multiply the two results:

LHS = (cos²θ/sinθ) × (sin²θ/cosθ) = sinθ · cosθ

Now simplify RHS: 1/(tanθ + cotθ):

tanθ + cotθ = sinθ/cosθ + cosθ/sinθ 

= (sin²θ + cos²θ)/(sinθ cosθ) = 1/(sinθ cosθ) 

So RHS = 1 ÷ [1/(sinθ cosθ)] = sinθ cosθ

LHS = sinθ cosθ = RHS 

Proved.

Question 4: Prove: (tanθ + secθ − 1) / (tanθ − secθ + 1) = (1 + sinθ) / cosθ

Solution: 

Take LHS = (tanθ + secθ − 1) / (tanθ − secθ + 1).

Replace 1 in numerator using sec²θ − tan²θ:

Numerator = tanθ + secθ − (sec²θ − tan²θ) 

= tanθ + secθ − (secθ − tanθ)(secθ + tanθ) 

= (secθ + tanθ)[1 − (secθ − tanθ)] 

= (secθ + tanθ)(1 − secθ + tanθ)

Denominator = tanθ − secθ + 1 = 1 − secθ + tanθ 

Cancel (1 − secθ + tanθ):

LHS = (secθ + tanθ)(1 − secθ + tanθ) / (1 − secθ + tanθ) 

= secθ + tanθ = 1/cosθ + sinθ/cosθ 

= (1 + sinθ)/cosθ = RHS 

LHS = (1 + sinθ)/cosθ = RHS. 

Proved.

Question 5: Evaluate sin 18° / cos 72° + √3 (tan 10° · tan 30° · tan 60° · tan 80°)

Solution:

For the first part: 18° + 72° = 90°, so sin 18° = cos 72°.

sin 18°/cos 72° = cos 72°/cos 72° = 1

For the second part: 10° + 80° = 90°, so tan 80° = cot 10°.

tan 10° × tan 80° = tan 10° × cot 10° = tan 10° × (1/tan 10°) = 1

Also, tan 30° = 1/√3, tan 60° = √3:

tan 30° × tan 60° = (1/√3)(√3) = 1

The product tan 10° × tan 30° × tan 60° × tan 80° = (tan 10° × tan 80°) × (tan 30° × tan 60°) = 1 × 1 = 1 

Full expression = 1 + √3 × 1 = 1 + √3

Question 6: If sin 3A = cos(A − 26°), where 3A is an acute angle, find the value of A.

Solution:

Since sin X = cos Y implies X + Y = 90° (for acute angles):

3A + (A − 26°) = 90° 

4A − 26° = 90° 

4A = 116° 

A = 29° 

Question 7: If x sin³θ + y cos³θ = sin θ cos θ, and x sin θ = y cos θ, prove that: x² + y² = 1

Solution:

From x sin θ = y cos θ:

x/y = cos θ/sin θ = cot θ 

So: x = y cos θ/sin θ

Substitute into x sin³θ + y cos³θ = sin θ cos θ:

(y cos θ/sin θ)(sin³θ) + y cos³θ = sin θ cos θ 

y cos θ sin²θ + y cos³θ = sin θ cos θ 

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

y cos θ (1) = sin θ cos θ y = sin θ

Then x = y cos θ/sin θ = sin θ · cos θ/sin θ = cos θ.

Therefore:

x² + y² = cos²θ + sin²θ = 1 

x² + y² = 1. 

Proved.

Question 8: If tan θ + sin θ = m and tan θ − sin θ = n, show that: m² − n² = 4√(mn)

Solution: 

m + n = 2 tan θ   and   m − n = 2 sin θ

m² − n² = (m+n)(m−n):

= (2 tan θ)(2 sin θ) = 4 tan θ sin θ

mn = (tan θ + sin θ)(tan θ − sin θ) = tan²θ − sin²θ:

= sin²θ/cos²θ − sin²θ = sin²θ(1/cos²θ − 1) = sin²θ(1 − cos²θ)/cos²θ = sin²θ · sin²θ/cos²θ = sin²θ tan²θ

Therefore 4√(mn):

4√(mn) = 4√(sin²θ tan²θ) = 4 sin θ tan θ

So m² − n² = 4 tan θ sin θ = 4 sin θ tan θ = 4√(mn) 

 

Questions PDF with worked-out examples for Class 10 Chapter 8: Introduction to Trigonometry, perfect for last-minute CBSE exam revision.

Class 10 Chapter 8: Introduction to Trigonometry HOTS PDF