Question

If $tan\Theta +\frac{1}{tan\Theta }=2,$ then the value of $ta{n}^2\Theta +\frac{1}{ta{n}^2\Theta }$

Answer 1

Given:- $\tan \theta +\frac{1}{\tan \theta }=2$

To find:- ${\tan }^2\theta +\frac{1}{{\tan }^2\theta }=?$

Now,

$\tan \theta +\frac{1}{\tan \theta }=2\left(\text{Given}\right)$

Squaring both sides, we have

${(\tan \theta +\frac{1}{\tan \theta })}^2={\left(2\right)}^2$

${\tan }^2\theta +\frac{1}{{\tan }^2\theta }+2\left(\tan \theta \right)\left(\frac{1}{\tan \theta }\right)=4(\because {(a+b)}^2=a^2+b^2+2ab)$

$\Rightarrow {\tan }^2\theta +\frac{1}{{\tan }^2\theta }+2=4$

$\Rightarrow {\tan }^2\theta +\frac{1}{{\tan }^2\theta }=4-2=2$

Thus ${\tan }^2\theta +\frac{1}{{\tan }^2\theta }=2$

Hence the correct answer is $2$.