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SSC CHSL 15 Nov 2025 Shift 1 Paper

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Question : 67 of 100

Marks: +1, -0

If

2Sin^{2}A+2\sqrt{2}CosA-3=0

, then find the value of tanA.

1
$\sqrt{2}$
$(1/\sqrt{2})$
0

Solution:

Concept:

Use the identity

\sin^{2}A = 1 - \cos^{2}A

to convert the equation into a quadratic in

\cos A

.

Explanation:

Start with

2\sin^{2}A + 2\sqrt{2}\cos A - 3 = 0

.

Substitute

\sin^{2}A = 1 - \cos^{2}A

:

2(1 - \cos^{2}A) + 2\sqrt{2}\cos A - 3 = 0

Simplify:

2 - 2\cos^{2}A + 2\sqrt{2}\cos A - 3 = 0

This gives:

-2\cos^{2}A + 2\sqrt{2}\cos A - 1 = 0

Multiply by

-1

:

2\cos^{2}A - 2\sqrt{2}\cos A + 1 = 0

Recognize it as a perfect square:

(\sqrt{2}\cos A - 1)^{2} = 0

So

\sqrt{2}\cos A = 1

, hence

\cos A = \frac{1}{\sqrt{2}}

.

Then

\sin A = \sqrt{1 - \cos^{2}A} = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}}

.

Thus

\tan A = \frac{\sin A}{\cos A} = \frac{1/\sqrt{2}}{1/\sqrt{2}} = 1

.

Answer:

A. 1

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