# Choose the correct option. Justify your choice.

(i) 9 sec^{2}A - 9 tan^{2}A =

(A) 1 (B) 9 (C) 8 (D) 0

(ii) (1 + tanθ + secθ) (1 + cotθ - cosecθ) =

(A) 0 (B) 1 (C) 2 (D) -1

(iii) (sec A + tan A) (1 - sin A) =

(A) sec A (B) sin A (C) cosec A (D) cos A

(iv) (1 + tan^{2} A)/(1 + cot^{2} A) =

(A) sec^{2}A (B) -1 (C) cot^{2}A (D) tan^{2}A

**Solution:**

We will use the basic trigonometric identities and properties of the trigonometric ratios to solve the problem.

sin^{2} A + cos^{2} A = 1

cosec^{2} A = 1 + cot^{2} A

sec^{2} A = 1 + tan^{2} A

(i) 9 sec^{2}A - 9 tan^{2}A

= 9 (sec^{2} A - tan^{2} A)

= 9 × 1 [By using the identity, 1 + sec^{2} A = tan^{2} A]

= 9

Thus, option (B) is the correct answer.

(ii) (1 + tanθ + secθ) (1 + cotθ - cosecθ) -------- (1)

We know that using the trigonometric ratios,

tan (x) = sin (x)/cos (x)

cot (x) = cos (x)/sin(x) = 1/tan (x)

sec (x) = 1/cos (x)

cosec (x) = 1/sin (x)

By substituting the above function in equation (1),

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

= [(cosθ + sinθ + 1)/cosθ] [(sinθ + cosθ - 1)/sinθ] (By taking LCM and multiplying)

= [(sinθ + cosθ)^{2} - (- 1)^{2}] / sinθ cosθ [Using a^{2} - b^{2} = (a + b)(a - b)]

= [sin^{2} θ + cos^{2} θ + 2 sinθ cosθ - 1] / sinθ cosθ

= (1 + 2 sinθ cosθ - 1) / sinθ cosθ (Using identity sin^{2} θ + cos^{2} θ = 1)

= 2 sinθ cosθ / sinθ cosθ

= 2

Hence, option (C) is correct.

(iii) (sec A + tan A) (1 - sin A)

We know that,

tan(x) = sin(x) / cos(x)

sec(x) = 1/cos(x)

By substituting these values in the given expression we get,

= [(1/cos A) + (sin A/cos A)] (1 - sin A)

= [(1 + sin A)/cos A] (1 - sin A)

= (1 - sin^{2} A) / cos A

= cos^{2} A/cos A (By using the identity sin^{2}θ + cos^{2}θ = 1)

= cos A

Hence, option (D) is correct.

(iv) (1 + tan^{2} A) / (1 + cot^{2} A)

We know that,

tan (x) = sin (x) / cos (x)

cot (x) = cos (x) / sin (x)

= 1/tan (x)

By substituting these in the given expression we get,

(1 + tan^{2} A) / (1 + cot^{2} A) = (1 + sin^{2} A/cos^{2} A) / (1 + cos^{2} A/sin^{2} A)

= [(cos^{2} A + sin^{2} A)/cos^{2} A] [(sin^{2} A + cos^{2} A)/sin^{2} A]

= (1/cos^{2} A)/(1/sin^{2} A) [By using the identity: sin^{2} A + cos^{2} A = 1]

= sin^{2} A/cos^{2} A

= tan^{2} A

Hence, option (D) is correct.

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 8

**Video Solution:**

## Choose the correct option. Justify your choice. (i) 9 sec²A - 9 tan²A = (A) 1 (B) 9 (C) 8 (D) 0 (ii) (1 + tanθ + secθ) (1 + cotθ - cosecθ) = (A) 0 (B) 1 (C) 2 (D) -1 (iii) (sec A + tan A) (1 - sin A) = (A) sec A (B) sin A (C) cosec A (D) cos A (iv) (1 + tan²A)/(1 + cot²A) = (A) sec²A (B) -1 (C) cot²A (D) tan²A

Maths NCERT Solutions Class 10 Chapter 8 Exercise 8.4 Question 4

**Summary:**

The correct answers for the following 9 sec^{2}A - 9 tan^{2}A, (1 + tan θ + sec θ)(1 + cot θ − cosec θ), (sec A + tan A) (1 -sin A) and (1 + tan^{2} A)/(1 + cot^{2} A) are: (B) 9, (C) 2, (D) cos A and (D) tan^{2} A, respectively.

**☛ Related Questions:**

- Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
- Write all the other trigonometric ratios of ∠A in terms of sec A.
- Evaluate:(i) (sin² 63° + sin² 27) / (cos² 17° + cos² 73°)(ii) sin 25° cos 65° + cos 25° sin 65°
- Prove the following identities, where the angles involved are acute angles for which the expressions are defined.(i) (cosecθ - cotθ)² = 1 - cosθ/1 + cosθ(ii) cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A(iii) tanθ/(1 - cotθ) + cotθ/(1 - tanθ) = 1 + secθ cosecθ(iv) (1 + sec A)/sec A = sin² A/(1 - cos A)(v) (cos A - sin A + 1)/cos A + sin A + 1 = cosec A + cot A(vi) 1 + sin A/(1 - sin A) = sec A+ tan A(vii) (sinθ - 2sin³ θ)/(2cosθ - cosθ) = tanθ(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A(ix) (cosec A - sin A)(sec A - cos A) = 1/(tan A + cot A)(x) [(1 + tan² A)/(1 + cot² A)] = [(1 - tan A/1 - cot A)²] = tan² A

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