SarkariPYQ

SSC CGL 2024 PYQ — Question 61

SSC CGL - 19 Sep 2024 - Shift 2 • Quantitative Aptitude • Q61

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The value of the expression 1−sin(2t)1+sin(2t)×cos(t)+sin(t)cos(t)−sin(t){1−sin(2t)\over 1+sin(2t)} ×{cos(t)+sin(t)\over cos(t)−sin(t)}is.

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Correct answer: B

Explanation

The correct answer is option 2 i.e. 1−tan(t)1+tan(t)1−tan(t)\over 1+tan(t).

1−sin(2t)1+sin(2t)×cos(t)+sin(t)cos(t)−sin(t){1−sin(2t)\over 1+sin(2t)} ×{cos(t)+sin(t)\over cos(t)−sin(t)}

=1−2sin⁡(t)cos⁡(t)1+2sin⁡(t)cos⁡(t)×cos⁡(t)+sin⁡(t)cos⁡(t)−sin⁡(t)=\frac{1-2\sin(t)\cos(t)}{1+2\sin(t)\cos(t)}\times\frac{\cos(t)+\sin(t)}{\cos(t)-\sin(t)}   [∵ sin2θ = 2 sinθ cosθ]

=sin⁡2(t)+cos⁡2(t)−2sin⁡(t)cos⁡(t)sin⁡2(t)+cos⁡2(t)+2sin⁡(t)cos⁡(t)×cos⁡(t)+sin⁡(t)cos⁡(t)−sin⁡(t)=\frac{\sin^2(t) + \cos^2(t) - 2\sin(t)\cos(t)}{\sin^2(t) + \cos^2(t) + 2\sin(t)\cos(t)}\times\frac{\cos(t)+\sin(t)}{\cos(t)-\sin(t)}

=(cos⁡(t)−sin⁡(t))2(cos⁡(t)+sin⁡(t))2×cos⁡(t)+sin⁡(t)cos⁡(t)−sin⁡(t)=\frac{(\cos(t) - \sin(t))^2}{(\cos(t) + \sin(t))^2}\times\frac{\cos(t)+\sin(t)}{\cos(t)-\sin(t)}

=cos⁡(t)−sin⁡(t)cos⁡(t)+sin⁡(t)=\frac{\cos(t) - \sin(t)}{\cos(t) + \sin(t)}

=1−tan⁡(t)1+tan(t)=\frac{1 - \tan(t)}{1 + tan(t)}

Therefore, the value of the expression is 1−tan(t)1+tan(t)1−tan(t)\over 1+tan(t).