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KCET 2014 Math Solved Paper

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Question : 36 of 60

Marks: +1, -0

If

\tan x = 3/4, \pi < x < \frac{3\pi}{2},

then the value of cos

\frac{x}{2}

is

$\frac{3}{\sqrt{10}}$
$-\frac{3}{\sqrt{10}}$
$-\frac{1}{\sqrt{10}}$
$\frac{1}{\sqrt{10}}$

Solution:

Given,

\tan x = 3/4

,

\pi < x < \frac{3\pi}{2}

\Rightarrow \frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}

(belongs to quadrant II)

In the second quadrant,

\cos\left(\frac{x}{2}\right)

is negative.

\sec^2 x = 1 + \tan^2 x

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

= \frac{16 + 9}{16} = \frac{25}{16}

\sec x = \pm \frac{5}{4}

Thus,

\cos x = -\frac{4}{5}

{since

\pi < x < \frac{3\pi}{2}

}

Using the formula,

\cos 2A = 2 \cos^2 A - 1

,

2 \cos^2 \left(\frac{x}{2}\right) = 1 + \cos x

= 1 + \left(-\frac{4}{5}\right)

2 \cos^2 \left(\frac{x}{2}\right) = \frac{1}{5}

\cos^2 \left(\frac{x}{2}\right) = \frac{1}{10}

\cos\left(\frac{x}{2}\right) = \pm \sqrt{\frac{1}{10}}

= -\frac{1}{\sqrt{10}}

{since

\frac{x}{2}

lies in the quadrant II}

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