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The value of $\frac{\cos \theta}{1+\sin \theta}$ is equal to
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2977 Upvotes
Verified Answer
The correct answer is:
$\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)$
$$
\begin{aligned}
\frac{\cos \theta}{1+\sin \theta} &=\frac{\sin \left(\frac{\pi}{2}-\theta\right)}{1+\cos \left(\frac{\pi}{2}-\theta\right)} \\
&=\frac{2 \sin \left(\frac{\pi}{4}-\frac{\theta}{2}\right) \cos \left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{2 \cos ^{2}\left(\frac{\pi}{4}-\frac{\theta}{2}\right)} \\
&=\frac{\sin \left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{\cos \left(\frac{\pi}{4}-\frac{\theta}{2}\right)}=\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)
\end{aligned}
$$
\begin{aligned}
\frac{\cos \theta}{1+\sin \theta} &=\frac{\sin \left(\frac{\pi}{2}-\theta\right)}{1+\cos \left(\frac{\pi}{2}-\theta\right)} \\
&=\frac{2 \sin \left(\frac{\pi}{4}-\frac{\theta}{2}\right) \cos \left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{2 \cos ^{2}\left(\frac{\pi}{4}-\frac{\theta}{2}\right)} \\
&=\frac{\sin \left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{\cos \left(\frac{\pi}{4}-\frac{\theta}{2}\right)}=\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)
\end{aligned}
$$
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