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SSC CGL 2026 PYQ — Question 22

SSC CGL Tier II - 19 Jan 2026 - Shift 1 • Quantitative Aptitude • Q22

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Simplify: tan⁡2θ−sin⁡2θ2+tan⁡2θ+cot⁡2θ\frac{\tan^2 \theta - \sin^2 \theta}{2 + \tan^2 \theta + \cot^2 \theta}

 

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Correct answer: D

Explanation

The correct answer is Option 4 i.e. sin⁡6θ\sin^6 \theta.

tan⁡2θ−sin⁡2θ2+tan⁡2θ+cot⁡2θ\frac{\tan^2 \theta - \sin^2 \theta}{2 + \tan^2 \theta + \cot^2 \theta}

=sin⁡2θcos⁡2θ−sin⁡2θ2+(sec⁡2θ−1)+(cosec2θ−1)=\frac{ \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta}{2 + (\sec^2 \theta - 1) + (cosec^2 \theta - 1)}

=sin⁡2θ(1cos⁡2θ−1)2+sec⁡2θ+cosec2θ−2=\frac{ \sin^2 \theta \left( \frac{1}{\cos^2 \theta} - 1 \right) }{2 + \sec^2 \theta + cosec^2 \theta - 2}

=sin⁡2θ⋅tan⁡2θsec⁡2θ+cosec2θ=\frac{ \sin^2 \theta \cdot \tan^2 \theta }{\sec^2 \theta + cosec^2 \theta}

=sin⁡4θcos⁡2θ1cos⁡2θ+1sin⁡2θ=\frac{\frac{\sin^4 \theta}{\cos^2 \theta} }{\frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}}

=sin⁡4θcos⁡2θsin⁡2θ+cos⁡2θsin⁡2θcos⁡2θ=\frac{\frac{\sin^4 \theta}{\cos^2 \theta} }{\frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}}

=sin⁡4θcos⁡2θ1sin⁡2θcos⁡2θ=\frac{\frac{\sin^4 \theta}{\cos^2 \theta} }{\frac{1}{\sin^2 \theta \cos^2 \theta}}

=sin⁡4θcos⁡2θ×sin⁡2θcos⁡2θ=\frac{\sin^4 \theta}{\cos^2 \theta} \times \sin^2 \theta \cos^2 \theta

=sin⁡6θ= \sin^6 \theta