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RRB Technician Grade I 20 Dec 2024 Shift 1 Paper

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Question : 98 of 100

Marks: +1, -0

If tan θ =

\frac{8}{15}

, then the value of

\frac{\sqrt{1-sin\theta }}{\sqrt{1+sin\theta }}

$\frac{3}{5}$
$\frac{3}{4}$
$\frac{5}{7}$
$\frac{3}{7}$

Solution:

Concept:

Use the given

\tan\theta

to find

\sin\theta

from a right triangle, then evaluate the given expression.

Explanation:

Given

\tan\theta = \frac{8}{15}

.

Let perpendicular

= 8

and base

= 15

.

Hypotenuse

= \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17

.

So

\sin\theta = \frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{8}{17}

.

Now substitute in the expression:

\frac{\sqrt{1-\sin\theta}}{\sqrt{1+\sin\theta}} = \sqrt{\frac{1-\frac{8}{17}}{1+\frac{8}{17}}}

= \sqrt{\frac{\frac{9}{17}}{\frac{25}{17}}} = \sqrt{\frac{9}{25}} = \frac{3}{5}

Thus, the required value is

\frac{3}{5}

.

Answer:

Option A:

\frac{3}{5}

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