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**Short Answer Type Questions**

**Q 1. Find the value of (sin ^{2}33° + sin^{2}57°) [Delhi 2019]**

⇒ sin

⇒ sin

⇒ 1 [Using sin

(i)

(ii) cos 48°

Ans:

= sin 42° - sin 42° = 0

**[CBSE 2019 (30/5/1)]Ans: **We have, sin x + cosy = 1

⇒ sin30° + cosy = 1

⇒ y = 60

Ans:

[CBSE 2019 (30/5/1)]

Ans:

**[CBSE 2019 (30/ 5/ 1)]****Ans: **Given: cos θ + sin θ = √2cos θ

Now, (cos θ + sin θ)^{2} + ( cos θ - sin θ)^{2}

cos^{2}θ + sin^{2}θ + 2cosθ. sin θ + cos^{2}θ + sin^{2}θ - 2cosθ. sinθ

1 + 1 = 2

Again, (cos θ + sin θ)^{2} + (cos θ - sin θ)^{2} = 2

⇒ (√2 cosθ)^{2} + (cos θ - sin θ)^{2} = 2

⇒ 2 cos^{2}θ + ( cos θ - sin θ)^{2} = 2

⇒ (cos θ - sin θ)^{2} = 2 - 2cos^{2}θ = 2 ( 1 - cos^{2}θ)

= 2sin^{2}θ

∴ cos θ - sin θ = √2 sin θ

Hence proved**Q 8. Prove that: ( sin A + cosec A) ^{2} + ( cos A + sec A)^{2} = 7 + tan^{2} A + cot^{2} 4 **

**[CBSE 2019 (30/1/2;]**

**Ans: **LHS = (sin A + cosec A)^{2} + (cos A + sec A)^{2}

=sin^{2} A + cosec^{2} A + 2sinA . cosec A + cos^{2} A + sec^{2} A + 2 cos A .sec A

= (sin^{2} A + cosec^{2} A + 2) + (cos^{2} A + sec^{2} A + 2) (sin A.cosec A = 1)

= (sin^{2} A + cos^{2}A) + (cosec^{2} A + sec^{2} A) + 4 (cos A. sec A = 1)

= 1 + 1+ cot^{2} A + 1 + tan^{2} A + 4

= 7 + tan^{2} A + cot^{2} A = RHS (∵ 1 + cot^{2} A = cosec^{2} A and 1 + tan^{2} A = sec^{2} A)**Q 9 . Evaluate:** ** [CBSE 2019 (30/5/2)]****Ans: **We have,**Q 10. If sin (A + 2B) and cos (A + 4B) = 0, A > B, and A + 4B ≤ 90 ^{0} then find A and B. [CBSE 2018 (C)]**

So, sin(A + 2B) = sin 60°

Hence A + 2B = 60° ..(i)

Also, we have

cos(A + 4B) = 0

cos(A + 4B) = cos 90°

A + 4B = 90° ..(ii)

Subtracting (ii) from (i), we have

-2B = - 30°

=> B = 15°

Put B = 15° in eq. (i), we have

A + 2(15°) = 60°

⇒ A + 30° = 60°

⇒ A = 30°

⇒ cot (90° - 2A) = cot (A - 18°)

[∵ cot (90° — 0) — tan θ]

⇒ 90° - 2A = A - 18°

⇒

∴ ∠A = 36°

**[CBSE, 2018 (C)]**

**Long Answer Type Questions**

**Q 1. Prove that: [CBSE 2019 (30/5/1)]****Ans: **LHS

**Q 2. Prove that:** ** [CBSE 2019 (30/5/1)]****Ans: **LHS

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