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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