SSC CHSL 10 Aug 2021 Shift 1 Paper

Section: Quantitative Aptitude

Question : 68 of 100

Marks: +1, -0

For

\theta: 0^{\circ} < \theta < 90^{\circ}

3 \sec \theta + 4 \cos \theta = 4 \sqrt{3}

(1 - \sin \theta + \cos \theta)

Solution:

Given,

0^{\circ} < \theta < 90^{\circ}

and

3 \sec \theta + 4 \cos \theta = 4 \sqrt{3}

\; \frac{3}{\cos \theta} + 4 \cos \theta = 4 \sqrt{3}

\left[ \because \sec \theta = \frac{1}{\cos \theta} \right]

\therefore 4 \cos^2 \theta - 4 \sqrt{3} \cos \theta + 3 = 0

\Rightarrow 4 \cos^2 \theta - 2 \sqrt{3} \cos \theta - 2 \sqrt{3} \cos \theta

\Rightarrow 2 \cos \theta (2 \cos \theta - \sqrt{3}) \; + 3 = 0

\; - \sqrt{3} (2 \cos \theta - \sqrt{3}) = 0

\Rightarrow (2 \cos \theta - \sqrt{3}) = 0

\therefore \cos \theta = \frac{\sqrt{3}}{2} = \cos 30^{\circ} \; [ \because 0^{\circ} < \theta < 90^{\circ} ]

\therefore

The value of

(1 - \sin \theta + \cos \theta) \; = 1 - \sin 30^{\circ} + \cos 30^{\circ}

\; = 1 - \frac{1}{2} + \frac{\sqrt{3}}{2}

\; = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2}

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