Prove that

Solution:

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, $\cos ^{2} \theta-\sin ^{2} \theta=\frac{1}{2}$.