If $ 3 \cos \theta=1 $, find the value of $ \frac{6 \sin ^{2} \theta+\tan ^{2} \theta}{4 \cos \theta} $

Given:

\( 3 \cos \theta=1 \)

To do:

We have to find the value of \( \frac{6 \sin ^{2} \theta+\tan ^{2} \theta}{4 \cos \theta} \).

Solution:  

$3 \cos \theta=\frac{1}{3}$

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

$=1-(\frac{1}{3})^2$

$=1-\frac{1}{9}$

$=\frac{9-1}{9}$

$=\frac{8}{9}$

$\tan ^{2} \theta=\frac{\sin^{2} \theta}{\cot ^{2} \theta}$

$=\frac{\frac{8}{9}}{(\frac{1}{3})^2}$

$=\frac{\frac{8}{9}}{\frac{1}{9}}$

$=8$

Therefore,

$\frac{6 \sin ^{2} \theta+\tan ^{2} \theta}{4 \cos \theta}=\frac{6(\frac{8}{9})+8}{4(\frac{1}{3})}$

$=\frac{\frac{2(8)}{3}+8}{\frac{4}{3}}$

$=\frac{\frac{16+3(8)}{3}}{\frac{4}{3}}$

$=\frac{40}{4}$

$=10$

The value of \( \frac{6 \sin ^{2} \theta+\tan ^{2} \theta}{4 \cos \theta} \) is $10$. 

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Updated on: 10-Oct-2022

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