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SSC CHSL 14 Nov 2025 Shift 3 Paper

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Question : 75 of 100

Marks: +1, -0

If

tan(x)+cot(x)=2

sin(x)cos(x)

.

1
$\frac{6}{7}$
$\frac{1}{2}$
$\frac{7}{9}$

Solution:

Concept:

Use the identity

tan(x)+cot(x)=\frac{1}{\sin(x)\cos(x)}

derived from writing both in terms of sine and cosine and applying

\sin^2(x)+\cos^2(x)=1

.

Explanation:

Given

tan(x)+cot(x)=2

.

Rewrite as

\frac{\sin(x)}{\cos(x)}+\frac{\cos(x)}{\sin(x)}=\frac{\sin^2(x)+\cos^2(x)}{\sin(x)\cos(x)}=\frac{1}{\sin(x)\cos(x)}

.

Therefore

\frac{1}{\sin(x)\cos(x)}=2

.

Cross-multiply:

\sin(x)\cos(x)=\frac{1}{2}

.

Answer:

\sin(x)\cos(x)=\frac{1}{2}

, which is option C.

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