Question

If $\theta \in (\frac{\pi }{4},\frac{\pi }{2})$ and $\sum _{n=1}^{\infty }\frac{1}{{\tan }^n\theta }=\sin \theta +\cos \theta$ then the value of $\tan \theta$ is

• A. $\sqrt{3}$
• B. $\sqrt{2}+1$
• C. $2+\sqrt{3}$
• D. $\sqrt{2}$

Answer 1

The correct option is A $\sqrt{3}$

$\sum _{n=1}^{\infty }\frac{1}{{\tan }^n\theta }=sin\theta +cos\theta$
$\frac{1}{\tan \theta }+\frac{1}{{\tan }^2\theta }+.......\infty =sin\theta +cos\theta$
$\frac{1}{\tan \theta }[1+\frac{1}{\tan \theta }+\frac{1}{{\tan }^2\theta }+...\infty ]=sin\theta +cos\theta$
$\frac{1}{\tan \theta }\left[\frac{1}{1-\frac{1}{\tan \theta }}\right]=sin\theta +cos\theta$
$\frac{1}{\tan \theta -1}=sin\theta +cos\theta$
$\frac{cos\theta }{sin\theta -cos\theta }=sin\theta +cos\theta$
$cos\theta ={\sin }^2\theta -{\cos }^2\theta$
$cos\theta =1-{\cos }^2\theta -{\cos }^2\theta$
$2{\cos }^2\theta +\cos \theta -1=0$
$2{\cos }^2\theta +2cos\theta -cos\theta -1=0$
$2cos\theta (cos\theta +1)-1(cos\theta +1)=0$
$(cos\theta +1)(2cos\theta -1)=0$
$cos\theta =-1,cos\theta =\frac{1}{2}$
$\theta =\pi$ (rejected)
$\because \theta ϵ(\frac{\pi }{4},\frac{\pi }{2})$
$\therefore cos\theta =\frac{1}{2}$
$\theta =60°$

$\therefore tan\theta =\sqrt{3}$