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UPSC NDA I 2017 Mathematics Solved Paper

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Question : 43 of 120

Marks: +1, -0

What is

\frac{1}{sin10^{∘}}-\frac{\sqrt{3}}{cos10^{∘}}

equal to?

0
1
2
4

Solution:

Concept:

Use trigonometric identities:

2\sin\theta\cos\theta = \sin 2\theta

and

\sin(x-y) = \sin x\cos y - \cos x\sin y

.

Explanation:

Start with the expression:

\frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ}

.

Take the LCM:

\frac{\cos 10^\circ - \sqrt{3}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}

.

Multiply numerator and denominator by 2:

\frac{2(\cos 10^\circ - \sqrt{3}\sin 10^\circ)}{2\sin 10^\circ \cos 10^\circ}

.

The denominator simplifies to

\sin 20^\circ

because

2\sin\theta\cos\theta = \sin 2\theta

.

Factor 4 from the numerator:

\frac{4\left(\frac{1}{2}\cos 10^\circ - \frac{\sqrt{3}}{2}\sin 10^\circ\right)}{\sin 20^\circ}

.

Note that

\frac{1}{2} = \sin 30^\circ

and

\frac{\sqrt{3}}{2} = \cos 30^\circ

.

So the numerator becomes

4(\sin 30^\circ \cos 10^\circ - \cos 30^\circ \sin 10^\circ)

.

Apply the identity

\sin(x-y) = \sin x\cos y - \cos x\sin y

: this equals

4\sin(30^\circ - 10^\circ) = 4\sin 20^\circ

.

Cancel

\sin 20^\circ

:

\frac{4\sin 20^\circ}{\sin 20^\circ} = 4

.

Answer:

4

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