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RRB Technician Grade 1 Model Paper 5

Question : 56 of 100

Marks: +1, -0

\sqrt{13}sin\theta =2

\frac{3tan\theta +\sqrt{13}sin\theta }{\sqrt{13}cos\theta -3tan\theta }

is:

Solution:

Concept:

Use the identity

\sin^2\theta + \cos^2\theta = 1

and the relation

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

to find all required trigonometric ratios and substitute them into the expression.

Explanation:

Given that

\sqrt{13}\sin\theta = 2

, we get

\sin\theta = \frac{2}{\sqrt{13}}

Using

\sin^2\theta + \cos^2\theta = 1

\cos^2\theta = 1 - \left(\frac{2}{\sqrt{13}}\right)^2

\cos^2\theta = 1 - \frac{4}{13} = \frac{9}{13}

Hence,

\cos\theta = \frac{3}{\sqrt{13}}

Now,

\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{2}{\sqrt{13}}}{\frac{3}{\sqrt{13}}} = \frac{2}{3}

Substitute these values into the given expression:

\frac{3\tan\theta + \sqrt{13}\sin\theta}{\sqrt{13}\cos\theta - 3\tan\theta}

= \frac{3 \times \frac{2}{3} + 2}{3 - 2}

= \frac{2 + 2}{1} = 4

Answer:

The value of the expression is

4

Therefore, the correct option is C.

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