Question

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

Answer 1

${\sum }_{n=1}^{\infty }\frac{1}{ta{n}^n\theta }=\frac{1}{tan\theta }+\frac{1}{ta{n}^2\theta }+\frac{1}{ta{n}^3\theta }+.......$

$=\frac{\frac{1}{tan\theta }}{1-\frac{1}{tan\theta }}$

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

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

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

$=sin\theta +cos\theta$ (given)

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

$\Rightarrow si{n}^2\theta -co{s}^2\theta =cos\theta$

$\Rightarrow 1-co{s}^2\theta -co{s}^2\theta =cos\theta$

$\Rightarrow 2co{s}^2\theta +cos\theta -1=0$

$\Rightarrow 2co{s}^2\theta +2cos\theta -cos\theta -1=0$

$\Rightarrow 2cos\theta (cos\theta +1)-1(cos\theta +1)=0$

$\Rightarrow (2cos\theta -1)(cos\theta +1)=0$

$\Rightarrow cos\theta =-1$ or $cos\theta =\frac{1}{2}$

if $cos\theta =-1\Rightarrow \theta =\pi$ but $\frac{\pi }{4}<\theta <\frac{\pi }{2}$ so $cos\theta \ne -1$

$\Rightarrow cos\theta =\frac{1}{2}$ since $\theta ϵ(\pi /4,\pi /2)$

$\Rightarrow \theta =\frac{\pi }{3}$

so $tan\theta =\sqrt{3}\Rightarrow ta{n}^2\theta =3$