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