Question:
If $3 \cos \theta=1$, find the value of $\frac{6 \sin ^{2} \theta+\tan ^{2} \theta}{4 \cos \theta}$
Solution:
Given: $3 \cos \theta=1$
We have to find the value of the expression $\frac{6 \sin ^{2} \theta+\tan ^{2} \theta}{4 \cos \theta}$
We have,
$3 \cos \theta=1$
$\Rightarrow \cos \theta=\frac{1}{3}$
$\sin \theta=\sqrt{1-\cos ^{2} \theta}=\sqrt{1-\left(\frac{1}{3}\right)^{2}}=\frac{\sqrt{8}}{3}$
$\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{\frac{\sqrt{8}}{3}}{\frac{1}{3}}=\sqrt{8}$
Therefore,
$\frac{6 \sin ^{2} \theta+\tan ^{2} \theta}{4 \cos \theta}=\frac{6 \times\left(\frac{\sqrt{8}}{3}\right)^{2}+(\sqrt{8})^{2}}{4 \times \frac{1}{3}}$
$=10$
Hence, the value of the expression is 10 .