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UPSC NDA I 2015 Mathematics Paper

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Question Numbers: 33-34

Direction: Let

\alpha

be the root of the equation

25 \cos^{2} \theta + 5 \cos \theta - 12 = 0

, where

\frac{\pi}{2} < \alpha < \pi

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Question : 33 of 120

Marks: +1, -0

What is tan α equal to?

$-3 / 4$
$3 / 4$
$-4 / 3$
$-4 / 5$

Solution:

Concept:

Solve the quadratic in

\cos \alpha

to find

\cos \alpha

, then use the quadrant to determine the sign of

\sin \alpha

, and finally compute

\tan \alpha = \frac{\sin \alpha}{\cos \alpha}

.

Explanation:

Given:

25 \cos^2 \alpha + 5 \cos \alpha - 12 = 0

and

\frac{\pi}{2} < \alpha < \pi

.

Treat as a quadratic in

\cos \alpha

with

a = 25

,

b = 5

,

c = -12

.

Apply the quadratic formula:

\cos \alpha = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 25 \cdot (-12)}}{2 \cdot 25} = \frac{-5 \pm \sqrt{25 + 1200}}{50} = \frac{-5 \pm 35}{50}

.

So

\cos \alpha = \frac{30}{50} = \frac{3}{5}

or

\cos \alpha = \frac{-40}{50} = -\frac{4}{5}

.

Since

\alpha

is in the second quadrant (

\frac{\pi}{2} < \alpha < \pi

),

\cos \alpha

is negative. Hence

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

.

In the second quadrant,

\sin \alpha

is positive.

\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}

.

Now

\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{3/5}{-4/5} = \frac{3}{-4} = -\frac{3}{4}

.

Answer:

\tan \alpha = -\frac{3}{4}

, which matches option A.

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