Question

If $\theta \in (\pi /4,\pi /2)$ 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

Answer: (A)

$\tan \theta >1\Rightarrow 0<\frac{1}{\tan \theta }<1$
$\text{ now, }\frac{1}{\tan \theta }+\frac{1}{{\tan }^2\theta }+\frac{1}{{\tan }^3\theta }+\dots \dots \infty =\sin \theta +\cos \theta$
$\frac{\frac{1}{\tan \theta }}{1-\frac{1}{\tan \theta }}=\sin \theta +\cos \theta \Rightarrow \frac{1}{\tan \theta -1}=\sin \theta +\cos \theta$
$\Rightarrow \frac{\cos \theta }{\sin \theta -\cos \theta }=\sin \theta +\cos \theta$
$\cos \theta ={\sin }^2\theta -{\cos }^2\theta =1-2{\cos }^2\theta$
$2{\cos }^2\theta +\cos \theta -1=0\Rightarrow (2\cos \theta -1)(\cos \theta +1)=0$
$\cos \theta =\frac{1}{2}\text{ or }\cos \theta =-1\text{ (rejected) }$
$\theta =\frac{\pi }{3}\Rightarrow \tan \theta =\sqrt{3}$