Question

Evaluate \[ \frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x} \]

• A. \(2\sec x\)
• B. \(2\cosec x\)
• C. \(\tan 2x\)
• D. \(\sin 2x\)

Hint:

Whenever an expression contains \(\sin x\) and \(1+\cos x\), try taking the LCM and use the identity \[ \sin^2x+\cos^2x=1 \] to simplify the numerator.

Answer 1

The Correct Option is B

Step 1: Take the LCM of the two fractions.
\[ \frac{\sin x}{1+\cos x} + \frac{1+\cos x}{\sin x} \] Taking the LCM, \[ = \frac{\sin^2x+(1+\cos x)^2} {\sin x(1+\cos x)} \]

Step 2: Simplify the numerator.
Using \[ \sin^2x+\cos^2x=1 \] we get \[ \sin^2x+(1+\cos x)^2 \] \[ = \sin^2x+1+2\cos x+\cos^2x \] \[ = (\sin^2x+\cos^2x)+1+2\cos x \] \[ = 1+1+2\cos x \] \[ = 2(1+\cos x) \]

Step 3: Cancel common factors.
Therefore, \[ \frac{2(1+\cos x)} {\sin x(1+\cos x)} = \frac{2}{\sin x} \] \[ = 2\cosec x \]

Step 4: Final conclusion.
Hence, \[ \boxed{2\cosec x} \] Therefore, the correct option is \[ \boxed{(2)\ 2\cosec x} \]