Introduction to Trigonometry l Exercise 8.3 l Class 10 l NCERT Solutions

Class 10 Maths (NCERT)

Chapter 8: Introduction to Trigonometry

Exercise 8.3 – Stepwise Solutions

Important Identities Used

• sin²θ + cos²θ = 1
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ
• tan θ = sin θ / cos θ,   cot θ = cos θ / sin θ
• sec θ = 1/cos θ,   cosec θ = 1/sin θ

1. Express sin A, sec A and tan A in terms of cot A.

Step 1: Use identity: 1 + cot²A = cosec²A

So, cosec A = √(1 + cot²A)   (A is acute ⇒ positive root)

Step 2: sin A = 1 / cosec A

sin A = 1 / √(1 + cot²A)

Step 3: tan A = 1 / cot A

tan A = 1/cot A

Step 4: sec A = √(1 + tan²A)

sec A = √(1 + (1/cot A)²)

sec A = √((cot²A + 1)/cot²A)

sec A = √(1 + cot²A) / cot A

Final:

• sin A = 1 / √(1 + cot²A)
• tan A = 1 / cot A
• sec A = √(1 + cot²A) / cot A

2. Write all other trigonometric ratios of ∠A in terms of sec A.

Step 1: sec A = 1/cos A ⇒ cos A = 1/sec A

Step 2: sin²A = 1 − cos²A

sin²A = 1 − (1/sec A)² = 1 − 1/sec²A

sin²A = (sec²A − 1)/sec²A

sin A = √(sec²A − 1) / sec A (acute angle)

Step 3: tan²A = sec²A − 1

tan A = √(sec²A − 1)

Step 4: cot A = 1/tan A

cot A = 1 / √(sec²A − 1)

Step 5: cosec A = 1/sin A

cosec A = sec A / √(sec²A − 1)

Final:

• cos A = 1/sec A
• sin A = √(sec²A − 1) / sec A
• tan A = √(sec²A − 1)
• cot A = 1 / √(sec²A − 1)
• cosec A = sec A / √(sec²A − 1)

3. Choose the correct option. Justify your choice.

(i) 9 sec²A − 9 tan²A = ?

Step 1: Factor 9

= 9(sec²A − tan²A)

Step 2: Use identity sec²A = 1 + tan²A

sec²A − tan²A = 1

So value = 9 × 1 = 9

Correct option: (B) 9

(ii) (1 + tan θ + sec θ)(1 + cot θ − cosec θ) = ?

Step 1: Rewrite each bracket using sin, cos.

1 + tan θ + sec θ = 1 + (sinθ/cosθ) + (1/cosθ)

= (cosθ + sinθ + 1)/cosθ

1 + cot θ − cosec θ = 1 + (cosθ/sinθ) − (1/sinθ)

= (sinθ + cosθ − 1)/sinθ

Step 2: Multiply:

= [(cosθ + sinθ + 1)(sinθ + cosθ − 1)] / (sinθ cosθ)

Step 3: Use (a+1)(a−1)=a²−1 with a = (sinθ + cosθ)

Numerator = (sinθ + cosθ)² − 1

= (sin²θ + cos²θ + 2sinθcosθ) − 1

= (1 + 2sinθcosθ) − 1 = 2sinθcosθ

Step 4: So value = (2sinθcosθ)/(sinθcosθ) = 2

Correct option: (C) 2

(iii) (sec A + tan A)(1 − sin A) = ?

Step 1: Write in sin and cos

sec A + tan A = (1/cosA) + (sinA/cosA) = (1 + sinA)/cosA

Step 2: Multiply:

( (1 + sinA)/cosA ) (1 − sinA)

= (1 − sin²A)/cosA

= (cos²A)/cosA

= cos A

Correct option: (D) cos A

(iv) (1 + tan²A) / (1 + cot²A) = ?

Step 1: Use identities

1 + tan²A = sec²A

1 + cot²A = cosec²A

Step 2: Ratio

= sec²A / cosec²A

= (1/cos²A) / (1/sin²A)

= sin²A / cos²A

= tan²A

Correct option: (D) tan²A

4. Prove the following identities (acute angles).

(i) (cosec θ − cot θ)² = (1 − cos θ)/(1 + cos θ)

Step 1: Start with LHS

(cosecθ − cotθ) = (1/sinθ) − (cosθ/sinθ) = (1 − cosθ)/sinθ

Step 2: Square

LHS = (1 − cosθ)²/sin²θ

Step 3: Use sin²θ = 1 − cos²θ = (1 − cosθ)(1 + cosθ)

LHS = (1 − cosθ)² / [(1 − cosθ)(1 + cosθ)]

LHS = (1 − cosθ)/(1 + cosθ) = RHS

Hence proved.

(ii) cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A

Step 1: Take LHS and make common denominator

LHS = [cos²A + (1 + sinA)²] / [cosA(1 + sinA)]

Step 2: Expand numerator

= [cos²A + (1 + 2sinA + sin²A)] / [cosA(1 + sinA)]

= [(sin²A + cos²A) + 1 + 2sinA] / [cosA(1 + sinA)]

= [1 + 1 + 2sinA] / [cosA(1 + sinA)]

= 2(1 + sinA)/[cosA(1 + sinA)]

= 2/cosA = 2 secA

Hence proved.

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

Step 1: Write tan and cot in sin, cos

tanθ/(1 − cotθ) = (sin/cos) / (1 − cos/sin)

= (sin/cos) / ((sin − cos)/sin)

= (sin/cos) × (sin/(sin − cos))

= sin²θ / [cosθ(sinθ − cosθ)]

cotθ/(1 − tanθ) = (cos/sin) / (1 − sin/cos)

= (cos/sin) / ((cos − sin)/cos)

= (cos/sin) × (cos/(cos − sin))

= cos²θ / [sinθ(cosθ − sinθ)]

= − cos²θ / [sinθ(sinθ − cosθ)]

Step 2: Add both terms

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

Step 3: Take common denominator sinθcosθ(sinθ − cosθ)

LHS = [ sin³θ − cos³θ ] / [ sinθcosθ(sinθ − cosθ) ]

Step 4: Use a³ − b³ = (a − b)(a² + ab + b²)

Numerator = (sinθ − cosθ)(sin²θ + sinθcosθ + cos²θ)

= (sinθ − cosθ)(1 + sinθcosθ)

Step 5: Cancel (sinθ − cosθ) with (sinθ − cosθ) (note: (sinθ − cosθ)=−(cosθ−sinθ), cancellation valid)

LHS = (1 + sinθcosθ)/(sinθcosθ)

= 1/(sinθcosθ) + 1

= 1 + (1/sinθ)(1/cosθ)

= 1 + cosecθ · secθ = RHS

Hence proved.

(iv) (1 + sec A)/sec A = sin²A/(1 − cos A)

Step 1: Simplify LHS

(1 + secA)/secA = 1/secA + 1 = cosA + 1

Step 2: Simplify RHS

RHS = sin²A/(1 − cosA)

= (1 − cos²A)/(1 − cosA)

= (1 − cosA)(1 + cosA)/(1 − cosA)

= 1 + cosA

LHS = RHS. Hence proved.

(v) (cos A − sin A + 1)/(cos A + sin A − 1) = cosec A + cot A

Step 1: Start with LHS and multiply numerator & denominator by (cosA + sinA + 1)

LHS = (cosA − sinA + 1)(cosA + sinA + 1) / [(cosA + sinA − 1)(cosA + sinA + 1)]

Step 2: Denominator is (x−1)(x+1)=x²−1 with x=cosA+sinA

Den = (cosA + sinA)² − 1

= (cos²A + sin²A + 2sinAcosA) − 1

= (1 + 2sinAcosA) − 1 = 2sinAcosA

Step 3: Numerator is ( (cosA+1) − sinA )((cosA+1) + sinA )

= (cosA + 1)² − sin²A

= (cos²A + 2cosA + 1) − sin²A

= (cos²A − sin²A) + 2cosA + 1

= (cos2A) + 2cosA + 1

Better simplify using sin²+cos²=1:

(cosA + 1)² − sin²A = cos²A + 2cosA + 1 − sin²A

= (cos²A − sin²A) + 2cosA + 1

= (cos²A − (1 − cos²A)) + 2cosA + 1

= (2cos²A − 1) + 2cosA + 1

= 2cosA(cosA + 1)

Step 4: So

LHS = [2cosA(cosA + 1)] / [2sinAcosA] = (cosA + 1)/sinA

= 1/sinA + cosA/sinA

= cosecA + cotA

Hence proved.

(vi) √((1 + sin A)/(1 − sin A)) = sec A + tan A

Step 1: Start with LHS

√((1 + sinA)/(1 − sinA))

Step 2: Multiply inside by (1 + sinA)/(1 + sinA)

= √( (1 + sinA)² / (1 − sin²A) )

Step 3: 1 − sin²A = cos²A

= √( (1 + sinA)² / cos²A )

Step 4: Take square root (acute ⇒ cosA>0)

= (1 + sinA)/cosA

= 1/cosA + sinA/cosA

= secA + tanA

Hence proved.

(vii) (sin θ − 2 sin³θ) / (2 cos³θ − cos θ) = tan θ

Step 1: Factor numerator and denominator

Numerator = sinθ(1 − 2sin²θ)

Denominator = cosθ(2cos²θ − 1)

Step 2: Use sin²θ = 1 − cos²θ

1 − 2sin²θ = 1 − 2(1 − cos²θ) = 1 − 2 + 2cos²θ = 2cos²θ − 1

Step 3: Substitute

LHS = sinθ(2cos²θ − 1) / [cosθ(2cos²θ − 1)]

= sinθ/cosθ = tanθ

Hence proved.

(viii) (sinA + cosecA)² + (cosA + secA)² = 7 + tan²A + cot²A

Step 1: Expand both squares

(sinA + cosecA)² = sin²A + cosec²A + 2(sinA·cosecA)

= sin²A + cosec²A + 2

(cosA + secA)² = cos²A + sec²A + 2(cosA·secA)

= cos²A + sec²A + 2

Step 2: Add

LHS = (sin²A + cos²A) + (cosec²A + sec²A) + (2 + 2)

= 1 + (cosec²A + sec²A) + 4

= 5 + cosec²A + sec²A

Step 3: Use identities

cosec²A = 1 + cot²A

sec²A = 1 + tan²A

LHS = 5 + (1 + cot²A) + (1 + tan²A)

= 7 + tan²A + cot²A = RHS

Hence proved.

(ix) (cosecA − sinA)(secA − cosA) = 1/(tanA + cotA)

Step 1: Simplify each bracket

cosecA − sinA = (1/sinA) − sinA = (1 − sin²A)/sinA = cos²A/sinA

secA − cosA = (1/cosA) − cosA = (1 − cos²A)/cosA = sin²A/cosA

Step 2: Multiply

LHS = (cos²A/sinA) (sin²A/cosA)

= (sinA cosA)

Step 3: Simplify RHS

tanA + cotA = (sinA/cosA) + (cosA/sinA)

= (sin²A + cos²A)/(sinA cosA)

= 1/(sinA cosA)

So, 1/(tanA + cotA) = sinA cosA

LHS = RHS. Hence proved.

(x) (1 + tan²A)/(1 + cot²A) = ((1 − tanA)/(1 − cotA))² = tan²A

Part 1: (1 + tan²A)/(1 + cot²A)

= sec²A / cosec²A

= (1/cos²A)/(1/sin²A)

= sin²A/cos²A

= tan²A

Part 2: ((1 − tanA)/(1 − cotA))²

Step 1: Write cotA = 1/tanA

1 − cotA = 1 − 1/tanA = (tanA − 1)/tanA

Step 2: Form the fraction

(1 − tanA)/(1 − cotA) = (1 − tanA) / ((tanA − 1)/tanA)

= (1 − tanA) × (tanA/(tanA − 1))

= (-(tanA − 1)) × (tanA/(tanA − 1))

= -tanA

Step 3: Square

((1 − tanA)/(1 − cotA))² = (-tanA)² = tan²A

Hence proved.

Final Answers (Quick)

1) sinA = 1/√(1+cot²A), secA = √(1+cot²A)/cotA, tanA = 1/cotA
2) cosA = 1/secA, sinA = √(sec²A−1)/secA, tanA = √(sec²A−1), cotA = 1/√(sec²A−1), cosecA = secA/√(sec²A−1)
3) (i) 9 (B), (ii) 2 (C), (iii) cosA (D), (iv) tan²A (D)
4) All identities (i) to (x) proved above.