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NIMCET 2016 Question Paper with solutions

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Question : 69 of 120

Marks: +1, -0

If cos θ =

\frac{5}{13}

\frac{3\pi}{2}

< θ < 2π , then

\tan 2\theta

is ;

$\frac{-120}{119}$
$\frac{-120}{169}$
$\frac{119}{169}$
$\frac{120}{119}$

Solution:

Concept:

\tan 2\theta = \frac{2\tan\theta}{1-\tan^{2}\theta}

\cos\theta = \frac{\text{base}}{\text{hypotenuse}}

\tan\theta = \frac{\text{perpendicular}}{\text{base}}

\text{Hypotenuse}^{2} = \text{Perpendicular}^{2} + \text{Base}^{2}

\cos\theta = \frac{5}{13}, \frac{3\pi}{2} < \theta < 2\pi

From above

\theta

lies in the fourth quadrant so the

\tan\theta

will be negative

Hypotenuse

=13

Base

=5

Perpendicular

= \sqrt{13^{2}+5^{2}} = \pm 12

\tan\theta = \frac{\text{perpendicular}}{\text{base}}

\tan\theta = \frac{-12}{5}, \tan\theta

is negative in the fourth quadrant

\tan 2\theta = \frac{ \frac{2 \times -12}{5} }{ 1 - \frac{-12^{2}}{52} }

\tan 2\theta = \frac{120}{119}

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