NIMCET 2015 — Mathematics PYQ

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Question

NIMCET 2015 — Mathematics PYQ

NIMCET | Mathematics | 2015

If $0 < x < \pi$ and $\cos x + \sin x = \frac{1}{2}$ , then the value of $\tan x$ is:

Choose the correct answer:

Explanation

\cos x + \sin x = \frac{1}{2}

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

\cos^2 x + \sin^2 x + 2 \sin x \cos x = \frac{1}{4}

1 + \sin 2x = \frac{1}{4}

\sin 2x = \frac{1}{4} - 1 = -\frac{3}{4}

\frac{2 \tan x}{1 + \tan^2 x} = -\frac{3}{4}

8 \tan x = -3 - 3 \tan^2 x

3 \tan^2 x + 8 \tan x + 3 = 0

\tan x = \frac{-8 \pm \sqrt{8^2 - 4(3)(3)}}{2(3)}

\tan x = \frac{-8 \pm \sqrt{64 - 36}}{6}

\tan x = \frac{-8 \pm \sqrt{28}}{6}

\tan x = \frac{-8 \pm 2\sqrt{7}}{6}

\tan x = \frac{-4 \pm \sqrt{7}}{3}

Since $0 < x < \pi$ and $\sin 2x < 0$ , $x$ is in the second quadrant.

In second quadrant, $\tan x$ is negative.

From options:

\tan x = \frac{-4 + \sqrt{7}}{3}

Correct Option: (b)