If $\sqrt{3} \tan \theta=1$ then evaluate $\left(\cos ^{2} \theta-\sin ^{2} \theta\right)$.
Given: $\sqrt{3} \tan \theta=1$
$\Rightarrow \tan \theta=\frac{1}{\sqrt{3}}$
Since, $\tan \theta=\frac{P}{B}$
$\Rightarrow P=1$ and $B=\sqrt{3}$
Using Pythagoras theorem,
$P^{2}+B^{2}=H^{2}$
$\Rightarrow H^{2}=1+3=4$
$\Rightarrow H=2$
Therefore,
$\sin \theta=\frac{P}{H}=\frac{1}{2}$
$\cos \theta=\frac{B}{H}=\frac{\sqrt{3}}{2}$
$\cos ^{2} \theta-\sin ^{2} \theta=\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}$
$=\frac{3}{4}-\frac{1}{4}=\frac{2}{4}$
$=\frac{1}{2}$
Hence, $\cos ^{2} \theta-\sin ^{2} \theta=\frac{1}{2}$.
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