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UPSC CDS 2 2021 Math Paper

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Question : 37 of 100

Marks: +1, -0

If

\frac{\cos^{2}\theta - 3\cos\theta + 2}{\sin^{2}\theta}=1

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

\sin^{2}\theta + \cos\theta

equal to

$\frac{5}{4}$
$\frac{3}{2}$
$\frac{7}{4}$
2

Solution: 👈: Video Solution

Given,

\frac{\cos^{2}\theta - 3\cos\theta + 2}{\sin^{2}\theta}=1

Where,

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

\Rightarrow \cos^{2}\theta - 3\cos\theta + 2 = \sin^{2}\theta

\Rightarrow \cos^{2}\theta - 3\cos\theta + 2 = 1 - \cos^{2}\theta

[\because \sin^{2}A + \cos^{2}A = 1]

\Rightarrow 2\cos^{2}\theta - 3\cos\theta + 1 = 0

\Rightarrow 2\cos^{2}\theta - 2\cos\theta - \cos\theta + 1 = 0

\Rightarrow 2\cos\theta(\cos\theta - 1) - 1(\cos\theta - 1) = 0

\Rightarrow (\cos\theta - 1)(2\cos\theta - 1) = 0

\Rightarrow \cos\theta = 1 \; \text{or} \; \cos\theta = \frac{1}{2}

\because

In the given interval

\theta \neq 0 \Rightarrow

\cos\theta \neq 1

\therefore \cos\theta = \frac{1}{2} \; \text{(only)}

\sin\theta = \sqrt{1 - \cos^{2}\theta}

= \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}

Now,

\sin^{2}\theta + \cos\theta

= \left(\frac{\sqrt{3}}{2}\right)^{2} + \frac{1}{2} = \frac{3}{4} + \frac{1}{2}

= \frac{3}{4} + \frac{2}{4} = \frac{5}{4}

Hence,

\sin^{2}\theta + \cos\theta = \frac{5}{4}

.

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