Simplify : 1+sin(x)/1-cos(x)+1-sin(x)/1+cos(x)

Correct Answer is : 2(1+sin(x)cos(x))sin2(x){2(1+sin(x)cos(x))}{sin²(x)}sin2(x)2(1+sin(x)cos(x))

Correct Answer is : 2(1+sin(x)cos(x))sin2(x){2(1+sin(x)cos(x))}{sin²(x)}sin2(x)2(1+sin(x)cos(x))

Test Index

SSC CHSL 14 Nov 2025 Shift 1 Paper

Question : 73 of 100

Marks: +1, -0

Simplify :

\frac{1+sin(x)}{1-cos(x)}+\frac{1-sin(x)}{1+cos(x)}

Solution:

Concept:

Use the identity

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

and rationalize each denominator to combine the fractions.

Explanation:

First term:

\frac{1+\sin x}{1-\cos x}

. Multiply numerator and denominator by

(1+\cos x)

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

Second term:

\frac{1-\sin x}{1+\cos x}

. Multiply numerator and denominator by

(1-\cos x)

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

Add the two terms:

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

Expand each product:

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

(1-\sin x)(1-\cos x) = 1 - \cos x - \sin x + \sin x \cos x

Sum:

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

Thus, expression becomes

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

Answer:

Option B:

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

Go to Question:

Prev Question

Next Question