Question

If ${(\tan \theta +\frac{1}{\cos \theta })}^2+{(\tan \theta -\frac{1}{\cos \theta })}^2=m\left(\frac{1+{\sin }^2\theta }{1-{\sin }^2\theta }\right)$.Find $m$

Answer 1

$\tan x=\frac{\sin x}{\cos x}$

So, LHS:-

Expand and put value

$a^2+b^2=(a+b{)}^2-2ab$

$\Rightarrow a=\tan \theta +\frac{1}{\cos \theta },b=\tan \theta -\frac{1}{\cos \theta }$

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

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

$\Rightarrow 4{\tan }^2\theta -2({\tan }^2\theta -\frac{1}{{\cos }^2\theta })$

$\Rightarrow 4{\tan }^2\theta -2{\tan }^2\theta +\frac{2}{{\cos }^2\theta }$

$\Rightarrow 2{\tan }^2\theta +\frac{2}{{\cos }^2\theta }$

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

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

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

$\Rightarrow m=2$.