Question

If $x=\frac{\sqrt{1-\sin 4\theta }+1}{\sqrt{1+\sin 4\theta }+1}$ then one of the values of $x$ is

• A. $-\tan \theta$
• B. $\cot \theta$
• C. $\tan (\frac{\pi }{4}+\theta )$
• D. $-\cot (\frac{\pi }{4}+\theta )$

Answer 1

Answer: (B), (C)

$\cot \theta$
$\tan (\frac{\pi }{4}+\theta )$

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

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

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

$=\tan \theta$ change

(or) [$\theta$ depends on the Quadrant]

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

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

$=\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }$ change it

$\Rightarrow \frac{1-\tan \theta }{1+\tan \theta }=\tan (\frac{\pi }{4}-\theta )$

$=+\cot (\frac{\pi }{4}+\theta )$