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RRB NTPC Mains or Stage 2 Model Paper 9 with solutions

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Question : 19 of 120

Marks: +1, -0

If

\frac{x \csc^2 30^{\circ} \cdot \sec^2 45^{\circ}}{8 \cos^2 45^{\circ} \cdot \sin^2 60^{\circ}} = \tan^2 60^{\circ} - \tan^2 30^{\circ}, \text{ then } x = \text{?}

1
– 1
2
0

Solution:

Concept:

Use standard trigonometric values and solve the resulting equation for

x

.

Explanation:

Given equation:

\frac{x \csc^2 30^\circ \sec^2 45^\circ}{8 \cos^2 45^\circ \sin^2 60^\circ}=\tan^2 60^\circ-\tan^2 30^\circ

Substitute:

\csc 30^\circ=2

,

\sec 45^\circ=\sqrt{2}

,

\cos 45^\circ=\frac{1}{\sqrt{2}}

,

\sin 60^\circ=\frac{\sqrt{3}}{2}

.

Numerator:

x(2^2)(\sqrt{2})^2=x \cdot 4 \cdot 2=8x

.

Denominator:

8\left(\frac{1}{\sqrt{2}}\right)^2\left(\frac{\sqrt{3}}{2}\right)^2=8 \cdot \frac{1}{2} \cdot \frac{3}{4}=3

.

So, LHS becomes

\frac{8x}{3}

.

RHS:

\tan^2 60^\circ-\tan^2 30^\circ=(\sqrt{3})^2-\left(\frac{1}{\sqrt{3}}\right)^2=3-\frac{1}{3}=\frac{8}{3}

.

Thus,

\frac{8x}{3}=\frac{8}{3}

.

Multiplying both sides by

3

:

8x=8

.

Therefore,

x=1

.

Answer:

The value of

x

is

1

.

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