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RRB NTPC Mains or Stage 2 Model Paper 3 with solutions

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Question : 5 of 120

Marks: +1, -0

If

2 - \cos^2

= 3 sin θcosθ, sin θ≠cosθ, then tan θis

$\frac{1}{2}$
0
$\frac{2}{3}$
$\frac{1}{3}$

Solution:

Concept:

The equation can be converted into a quadratic in

\tan\theta

using

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

.

Explanation:

Given:

2-\cos^2\theta=3\sin\theta\cos\theta

.

Here

\cos\theta\neq0

, so divide throughout by

\cos^2\theta

:

2\sec^2\theta-1=3\tan\theta

Use

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

:

2(1+\tan^2\theta)-1=3\tan\theta

Simplify:

2\tan^2\theta-3\tan\theta+1=0

Factorise:

(2\tan\theta-1)(\tan\theta-1)=0

So,

\tan\theta=\frac{1}{2}

or

\tan\theta=1

.

Since

\sin\theta\neq\cos\theta

, we get

\tan\theta\neq1

.

Hence,

\tan\theta=\frac{1}{2}

.

Answer:

\frac{1}{2}

(Option A)

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