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The value of $\large\frac{\cos\theta}{1+\sin\theta}$ is equal to

$\begin{array}{1 1}(a)\;\tan\big(\large\frac{\theta}{2}-\frac{\pi}{4}\big)&(b)\;\tan \big(-\large\frac{\pi}{4}-\frac{\theta}{2}\big)\\(c)\;\tan\big(\large\frac{\pi}{4}-\frac{\theta}{2}\big)&(d)\;\tan\big(\large\frac{\pi}{4}+\frac{\theta}{2}\big)\end {array}$

1 Answer

We have $\large\frac{\cos\theta}{1+\sin\theta}$
$\Rightarrow \large\frac{\sin(\Large\frac{\pi}{2}-\normalsize \theta)}{1+\cos(\Large\frac{\pi}{2}-\normalsize\theta)}$
$\Rightarrow \large\frac{2\sin(\Large\frac{\pi}{4}-\frac{\theta}{2})\cos(\Large\frac{\pi}{4}-\frac{\theta}{2})}{2\cos^2(\Large\frac{\pi}{4}-\frac{\theta}{2})}$
$\Rightarrow \tan\big(\large\frac{\pi}{4}-\frac{\theta}{2})$
Hence (c) is the correct option.
answered Oct 4, 2013 by sreemathi.v
 
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