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RRB NTPC Mains or Stage 2 Model Paper 7 with solutions

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Question : 18 of 120

Marks: +1, -0

If tan q = 1, then find the value of

\frac{8\sin\theta +5\sin\theta}{\sin^3\theta - 2\cos^3\theta + 7\cos\theta}

2
$2\frac{1}{2}$
3
$\frac{4}{5}$

Solution:

Concept:

If

\tan\theta=1

, then

\sin\theta=\cos\theta

and

\sin^2\theta=\frac{1}{2}

.

Explanation:

Given:

\tan\theta=1

.

So,

\sin\theta=\cos\theta

.

Numerator:

8\sin\theta+5\sin\theta=13\sin\theta

.

Denominator:

\sin^3\theta-2\cos^3\theta+7\cos\theta

.

Replace

\cos\theta

by

\sin\theta

:

=\sin^3\theta-2\sin^3\theta+7\sin\theta

.

=-\sin^3\theta+7\sin\theta

.

=\sin\theta(7-\sin^2\theta)

.

Now the expression becomes:

\frac{13\sin\theta}{\sin\theta(7-\sin^2\theta)}=\frac{13}{7-\sin^2\theta}

.

Since

\tan\theta=1

,

\theta=45^\circ

, so

\sin^2\theta=\frac{1}{2}

.

Thus, value

=\frac{13}{7-\frac{1}{2}}=\frac{13}{\frac{13}{2}}=2

.

Answer:

2

, which is option A.

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