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Trigonometry Practice Test 1

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Question : 10 of 40

Marks: +1, -0

The value of

(\sqrt{3}\csc 20^{\circ} - \sec 20^{\circ})

is:

2
$2 \sqrt{3}$
4
$\frac{\sqrt{8}}{2}$

Solution: 👈: Video Solution

Concept:

Simplify the trigonometric expression using standard identities.

Explanation:

Write

\sqrt{3}

as

\tan 60^\circ

.

So,

\sqrt{3}\csc 20^\circ - \sec 20^\circ = \tan 60^\circ \csc 20^\circ - \sec 20^\circ

.

Convert into sine and cosine:

= \frac{\sin 60^\circ}{\cos 60^\circ}\cdot\frac{1}{\sin 20^\circ} - \frac{1}{\cos 20^\circ}

= \frac{\sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ}{\cos 60^\circ \sin 20^\circ \cos 20^\circ}

Use

\sin A \cos B - \cos A \sin B = \sin(A-B)

:

= \frac{\sin(60^\circ - 20^\circ)}{\cos 60^\circ \sin 20^\circ \cos 20^\circ}

= \frac{\sin 40^\circ}{\frac{1}{2}\sin 20^\circ \cos 20^\circ}

Since

2\sin 20^\circ \cos 20^\circ = \sin 40^\circ

, we get

\sin 20^\circ \cos 20^\circ = \frac{1}{2}\sin 40^\circ

.

= \frac{\sin 40^\circ}{\frac{1}{2}\cdot \frac{1}{2}\sin 40^\circ}

= \frac{\sin 40^\circ}{\frac{1}{4}\sin 40^\circ} = 4

Answer:

The value is

4

.

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