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Question: Answered & Verified by Expert
$\frac{1+\cos \theta-\sin \theta}{1+\cos \theta+\sin \theta}+\frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}=$
MathematicsTrigonometric EquationsAP EAMCETAP EAMCET 2023 (18 May Shift 1)
Options:
  • A $2 \sec \theta$
  • B $2 \operatorname{cosec} \theta$
  • C $2 \tan \theta$
  • D $2 \cot \theta$
Solution:
2384 Upvotes Verified Answer
The correct answer is: $2 \sec \theta$
$1+\cos \theta-\sin \theta$
$\begin{aligned} & =\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)^2+\left(\cos ^2 \frac{\theta}{2}-\sin ^2 \frac{\theta}{2}\right) \\ & =\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}+\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right) \\ & =2 \cos \frac{\theta}{2}\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)\end{aligned}$
Similarly, $1+\cos \theta+\sin \theta$ $=2 \cos \frac{\theta}{2}\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)$
Now, $\frac{1+\cos \theta-\sin \theta}{1+\cos \theta+\sin \theta}+\frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}$
$\begin{aligned} & =\frac{2 \cos \frac{\theta}{2}\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)}{2 \cos \frac{\theta}{2}\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)}+\frac{2 \cos \frac{\theta}{2}\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)}{2 \cos \frac{\theta}{2}\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)} \\ & =\frac{\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)^2+\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)^2}{\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)}\end{aligned}$
$\begin{aligned} & =\frac{1-\sin \theta+1+\sin \theta}{\cos 2 \theta}=\frac{2}{\cos 2 \theta} \\ & =2 \sec 2 \theta\end{aligned}$

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