If $ 2 \sin ^{2} \theta-\cos ^{2} \theta=2 $, then find the value of $ \theta $.

Given:

\( 2 \sin ^{2} \theta-\cos ^{2} \theta=2 \)

To do:

We have to find the value of \( \theta \).

Solution:  

$2 \sin ^{2} \theta-\cos ^{2} \theta=2$

$\Rightarrow 2 \sin ^{2} \theta-(1-\sin ^{2} \theta)=2$        [Since $\sin ^{2} \theta+\cos ^{2} \theta=1$]

$\Rightarrow 2 \sin ^{2} \theta+\sin ^{2} \theta=2+1$

$\Rightarrow 3 \sin ^{2} \theta=3$

$\Rightarrow \sin ^{2} \theta=1$

$\Rightarrow \sin \theta=1$

$\Rightarrow \sin \theta=\sin 90^{\circ}$         [Since $\sin 90^{\circ}=1$]

Comparing on both sides, we get,

$\theta=90^{\circ}$

The value of $\theta$ is $90^{\circ}$. 

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

105 Views

Kickstart Your Career

Get certified by completing the course

Get Started