Question

If $tan\alpha =\frac{1+\sqrt{1+\sin 2\theta }}{1-\sqrt{1-\sin 2\theta }}$ where $0<\theta <\frac{\pi }{4}$ then value of $\alpha$ can be

• A. $\frac{\pi }{2}-\frac{\theta }{2}$
• B. $\frac{\pi }{4}+\frac{\theta }{2}$
• C. $\pi +\theta$
• D. $\theta$

Answer 1

The correct option is B $\frac{\pi }{4}+\frac{\theta }{2}$

$\tan \alpha =\frac{1+\sqrt{1+\sin 2\theta }}{1-\sqrt{1-\sin 2\theta }}$

$=\frac{1+\sqrt{{\sin }^2\theta +{\cos }^2\theta +\sin 2\theta }}{1-\sqrt{{\sin }^2\theta +{\cos }^2\theta -\sin 2\theta }}$ by replacing $1={\sin }^2\theta +{\cos }^2\theta$

$=\frac{1+\sqrt{{\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta }}{1-\sqrt{{\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta }}$ by using multiple angle formula $\sin 2\theta =2\sin \theta \cos \theta$

$=\frac{1+\sqrt{{(\sin \theta +\cos \theta )}^2}}{1-\sqrt{{(\sin \theta -\cos \theta )}^2}}$

$=\frac{1+\sin \theta +\cos \theta }{1-\sin \theta +\cos \theta }$

$=\frac{(1+\cos \theta )+\sin \theta }{(1+\cos \theta )-\sin \theta }$

$=\frac{(1+\cos \theta )+\sin \theta }{(1+\cos \theta )-\sin \theta }\times \frac{(1+\cos \theta )+\sin \theta }{(1+\cos \theta )+\sin \theta }$ by rationalising the denominator.

$=\frac{{(1+\cos \theta )}^2+{\sin }^2\theta +2\sin \theta (1+\cos \theta )}{{(1+\cos \theta )}^2-{\sin }^2\theta }$

$=\frac{1+({\cos }^2\theta +{\sin }^2\theta +2(\sin \theta +\cos \theta +\sin \theta \cos \theta ))}{1+{\cos }^2\theta +2\cos \theta -{\sin }^2\theta }$

$=\frac{2+2\sin \theta +2\cos \theta +2\sin \theta \cos \theta }{2{\cos }^2\theta +2\cos \theta }$ on simplification

$=\frac{2[(1+\sin \theta )+\cos \theta (1+\sin \theta )]}{2\cos \theta (1+\cos \theta )}$

$=\frac{1+\sin \theta }{\cos \theta }$

$=\frac{1+\sin \theta }{\cos \theta }\times \frac{1-\sin \theta }{1-\sin \theta }$ by rationalising the numerator

$=\frac{1-{\sin }^2\theta }{\cos \theta (1-\sin \theta )}$

$=\frac{{\cos }^2\theta }{\cos \theta (1-\sin \theta )}$

$=\frac{\cos \theta }{(1-\sin \theta )}$

$=\frac{\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}}{1-\frac{2\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}}$

$=\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}-2\tan \frac{\theta }{2}}$

$=\frac{(1-\tan \frac{\theta }{2})(1+\tan \frac{\theta }{2})}{{(1-\tan \frac{\theta }{2})}^2}$

$=\frac{(1+\tan \frac{\theta }{2})}{(1-\tan \frac{\theta }{2})}$

$=\frac{(\tan \frac{\pi }{4}+\tan \frac{\theta }{2})}{(1-\tan \frac{\pi }{4}\tan \frac{\theta }{2})}$ by taking $1=\tan \frac{\pi }{4}$

$\Rightarrow \tan \alpha =\tan (\frac{\pi }{4}+\frac{\theta }{2})$

$\therefore \alpha =\frac{\pi }{4}+\frac{\theta }{2}$