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SSC CPO 2023 PYQ — Question 117

SSC CPO - 3 Oct 2023 - Shift 1 • Quantitative Aptitude • Q117

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if cos⁡θ−sin⁡θ=2sin⁡θ,\cos θ - \sin θ = \sqrt 2 \sin θ, then (cos θ + sinθ) is: 

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Correct answer: B

Explanation

The correct answer is Option 2 i.e. 2cosθ\sqrt2 cos θ​​​​​​​​​​​​​​​​​​​​​.

cosθ−sinθ=2sinθcos θ - sin θ = \sqrt 2 sin θ

Squaring both sides

⇒ (cosθ−sinθ)2=(2sinθ)2(cos θ - sin θ)^2 = {(\sqrt 2 sin θ})^2

⇒ (cos2θ+sin2θ−2cosθsinθ)=2sin2θ(cos ^2θ + sin ^2θ-2cosθsinθ) = {2 sin^2θ}

⇒ 1−2cosθsinθ=2sin2θ1-2cosθsinθ = {2 sin^2θ}

⇒ 1−2sin2θ=2sinθcosθ1-2sin^2θ = {2 sinθcosθ}......(1)

Solving 

(cosθ+sinθ) (cos θ + sinθ)

On squaring

⇒ (cosθ+sinθ)2​​=cos2θ+sin2θ+2cosθsinθ (cos θ + sinθ) ^2​​ = cos^2θ+sin^2θ+2cosθsinθ

⇒ (cosθ+sinθ)2= (cos θ + sinθ) ^2 =1+2cosθsinθ1+2cosθsinθ

On putting the value of 2 sinθ cosθ

⇒ (cosθ+sinθ)2= (cos θ + sinθ) ^2=1+1−2sin2θ1+1 - 2sin^2θ = 2−2sin2θ2 - 2sin^2θ

⇒ (cosθ+sinθ)2= (cos θ + sinθ) ^2= 2(1−sin2θ)2(1 - sin^2θ)

⇒ (cosθ+sinθ)2=2cos2θ (cos θ + sinθ) ^2 = 2cos^2θ

⇒ (cosθ+sinθ)=2cosθ (cos θ + sinθ) = \sqrt2 cosθ