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SSC CPO 2022 PYQ — Question 119

SSC CPO - 9 Nov 2022 - Shift 1 • Quantitative Aptitude • Q119

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If x=2sin⁡θ(1+cos⁡θ+sin⁡θ),x = \frac {2 \sin \theta}{(1 + \cos \theta + \sin \theta)}, then the value of 1−cos⁡θ+sin⁡θ(1+sin⁡θ)\frac {1 - \cos \theta + \sin \theta}{(1 + \sin \theta)} is:

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Correct answer: B

Explanation

The correct answer is option 2 i.e. x.

 If x=2sin⁡θ(1+cos⁡θ+sin⁡θ)x = \frac {2 \sin \theta}{(1 + \cos \theta + \sin \theta)}

⇒ 2sin⁡θ(1−cos⁡θ+sin⁡θ)(1+cos⁡θ+sin⁡θ)(1−cos⁡θ+sin⁡θ) \frac {2 \sin \theta (1 - \cos \theta + \sin \theta)}{(1 + \cos \theta + \sin \theta)(1 - \cos \theta + \sin \theta)} = x [Rationalizing the denominator]

⇒ 2sin⁡θ(1−cos⁡θ+sin⁡θ)(1+sin⁡θ)2−(cos⁡2θ) \frac {2 \sin \theta (1 - \cos \theta + \sin \theta)}{(1 + \sin \theta)^2- (\cos^2 \theta)} = x

⇒ 2sin⁡θ−2cos⁡θsin⁡θ+2sin⁡2θ(1+sin⁡2θ+2sin⁡θ−cos⁡2θ) \frac {2 \sin \theta - 2\cos \theta\sin \theta+ 2\sin^2 \theta}{(1 + \sin^2 \theta +2 \sin \theta - \cos^2 \theta)} = x

⇒ 2sinθ(1+cosθ−sinθ)/2sin2θ+2sinθ = x

⇒2sinθ(1+cosθ−sinθ)/2sinθ(1+sinθ) = x

⇒ (1+ cosθ − sinθ)/(1+sinθ) = x