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RRB NTPC Mains or Stage 2 Model Paper 3 with solutions

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Question : 4 of 120

Marks: +1, -0

I f

4x = \sec\theta \text{ and } \frac{4}{X} = \tan\theta

8 \left[ x^2 - \frac{1}{x^2} \right]

is

$\frac{1}{16}$
$\frac{1}{8}$
$\frac{1}{2}$
$\frac{1}{4}$

Solution:

Concept:

Use the identity

\sec^2\theta - \tan^2\theta = 1

.

Explanation:

Given

4x=\sec\theta

, so

x=\frac{\sec\theta}{4}

.

Also

\frac{4}{x}=\tan\theta

, so

\frac{1}{x}=\frac{\tan\theta}{4}

.

Squaring gives

x^2=\frac{\sec^2\theta}{16}

and

\frac{1}{x^2}=\frac{\tan^2\theta}{16}

.

Then

8\left(x^2-\frac{1}{x^2}\right)=8\left(\frac{\sec^2\theta-\tan^2\theta}{16}\right)

.

Using

\sec^2\theta-\tan^2\theta=1

, this becomes

8\left(\frac{1}{16}\right)

.

So the value is

\frac{1}{2}

.

Answer:

\frac{1}{2}

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