### Question 13

SSC-CGL 2018 June 4 Shift 2

If $\displaystyle \theta$ is acute and

$\displaystyle \cos^2 \theta = 3 (\cot^2 \theta - \cos^2 \theta)$, then the value of

$\displaystyle \bigg(\frac12 \sec \theta + \sin \theta\bigg)^{-1}$ is?

- $\displaystyle 2(\sqrt 3 - 1)$
- $\displaystyle \sqrt 3 + 1$
- $\displaystyle \sqrt 3 + 2$
- $\displaystyle 2(2 - \sqrt 3)$

### Solution in Short

First use the given expression to obtain the value of θ = 60°.

Then put sec 60° = 2, and sin 60° = √3/2 in the required expression. You will have to take the reciprocal and then rationalize to obtain 2 (2 - √3) as the answer!

### Solution in Detail

$\displaystyle \cos^2 \theta = 3 (\cot^2 \theta - \cos^2 \theta)$

$\displaystyle \therefore \cos^2 \theta = 3 \bigg(\frac{1}{\tan^2 \theta} - \cos^2 \theta \bigg)$

$\displaystyle \therefore \cos^2 \theta = 3 \bigg(\frac{1 - \cos^2 \theta \times {\tan^2 \theta}}{\tan^2 \theta} \bigg)$

$\displaystyle \therefore \cos^2 \theta = 3 \bigg(\frac{1 - \sin^2 \theta}{\tan^2 \theta} \bigg)$

$\displaystyle \therefore \cos^2 \theta = 3 \bigg(\frac{\cos^2 \theta}{\tan^2 \theta} \bigg)$

$\displaystyle \therefore \cancel{\cos^2 \theta} = 3 \bigg(\frac{\cancel{\cos^2 \theta}}{\tan^2 \theta} \bigg)$

$\displaystyle \therefore \tan^2 \theta = 3$

$\displaystyle \therefore \tan \theta = \sqrt 3$

only +ve root taken because theta is acute.

$\displaystyle \therefore \theta = 60 \degree$

Now $\displaystyle \sec 60 \degree = 2$

and $\displaystyle \sin 60 \degree = \frac{\sqrt 3}{2}$

$\displaystyle \therefore \bigg(\frac12 \sec \theta + \sin \theta\bigg)^{-1}$

$\displaystyle = \bigg(\frac 22 + \frac{\sqrt 3}{2}\bigg)^{-1}$

$\displaystyle = \frac{2}{2 + \sqrt 3}$

$\displaystyle = \frac{2}{2 + \sqrt 3} \times \frac{2 - \sqrt 3}{2 - \sqrt 3}$

$\displaystyle = 2(2 - \sqrt 3) \:\underline{Ans}$

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