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SSC CGL 2020 PYQ — Question 51

SSC CGL Tier II - Quant - 16 Nov 2020 • Quantitative Aptitude • Q51

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The value of (sinθ+cosθ−1)(sinθ−cosθ+1)\frac{(sinθ + cosθ - 1)}{(sinθ - cosθ + 1)} × 1+sinθ1−sinθ\sqrt{\frac{1 + sin \theta}{1 - sin \theta}}  is:

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Correct answer: D

Explanation

The correct answer is option 4 i.e. 1.

= sinθ+cosθ−1sinθ−cosθ+1×1+sinθ1−sinθ\frac{sin \theta + cos \theta -1}{sin \theta - cos \theta +1} \times \sqrt{\frac{1 + sin \theta}{1-sin \theta}}

on rationalization we get,

= (sinθ+cosθ−1)(sinθ−cosθ+1)\frac{(sinθ + cosθ - 1)}{(sinθ - cosθ + 1)}× (sinθ−cosθ−1)(sinθ−cosθ−1)\frac{(sinθ - cosθ - 1)}{(sinθ - cosθ - 1)} 1+sinθ1−sinθ\sqrt{\frac{1 + sin \theta}{1 - sin \theta}} ×1+sinθ1+sinθ\sqrt{\frac{1 + sin \theta}{1 + sin \theta}}

= sin2θ−sinθcosθ−sinθ+sinθcosθ−cos2θ−cosθ−sinθ+cosθ+1sin2θ+cos2θ−2sinθcosθ−1\frac{sin^2θ- sinθcosθ - sinθ + sinθcosθ - cos^2θ - cosθ - sinθ + cosθ + 1}{sin^2θ + cos^2θ - 2sinθcosθ - 1 }× (1+sinθ)(cosθ)\frac{(1 + sinθ)}{(cosθ)}

=  sin2θ−cos2θ−2sinθ+1−2sinθcos\frac{sin^2θ - cos^2θ - 2 sinθ + 1}{-2sinθcos }×(1+sinθ)(cosθ)\frac{(1 + sinθ)}{(cosθ)}

=  sin2θ−cos2θ−2sinθ+sin2θ+cos2θ−2sinθcos\frac{sin^2θ - cos^2θ - 2 sinθ + sin^2θ + cos^2θ }{-2sinθcos }×(1+sinθ)(cosθ)\frac{(1 + sinθ)}{(cosθ)}

=  =  2sin2θ−2sinθ−2sinθcos\frac{2sin^2θ - 2 sinθ }{-2sinθcos }×(1+sinθ)(cosθ)\frac{(1 + sinθ)}{(cosθ)}

= 2sinθ(sinθ−1)(−2sinθcosθ)\frac{2sinθ(sinθ - 1)}{( -2sinθcosθ ) }× (1+sinθ)(cosθ)\frac{(1 + sinθ)}{(cosθ)}

= (sin2θ - 1)/(-cos2θ )

= -cos2θ/-cos2θ

= 1