If 3cosθ = 1, find the value of 6 sin2 θ+tan2 θ4 cos θ

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 .

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now