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Question

The value of \(\frac{{\left( {\sin \theta - cos\theta } \right)\left( {1\; + \;\tan \theta \; + \;cot\theta } \right)}}{{1\; + \;sin\theta cos\theta }}\)

This question was previously asked in
SSC CGL 2018 (Tier 2) Statistics Previous Year Paper (22-feb-2018)
The correct answer is

secθ – cosecθ

Step 1 — Simplify the bracket \(1+\tan\theta+\cot\theta\):

\[1+\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}=\frac{\sin\theta\cos\theta+\sin^{2}\theta+\cos^{2}\theta}{\sin\theta\cos\theta}=\frac{1+\sin\theta\cos\theta}{\sin\theta\cos\theta}\]

Step 2 — Substitute back and cancel:

\[\frac{(\sin\theta-\cos\theta)\cdot\dfrac{1+\sin\theta\cos\theta}{\sin\theta\cos\theta}}{1+\sin\theta\cos\theta}=\frac{\sin\theta-\cos\theta}{\sin\theta\cos\theta}\]

Step 3 — Split the fraction:

\[\frac{\sin\theta}{\sin\theta\cos\theta}-\frac{\cos\theta}{\sin\theta\cos\theta}=\frac{1}{\cos\theta}-\frac{1}{\sin\theta}=\sec\theta-\csc\theta\]

Therefore the expression equals \(\sec\theta-\csc\theta\).

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