Question

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

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

Hint:

When adding algebraic fractions, find a common denominator.
Use fundamental trigonometric identities like \(\sin^2\theta + \cos^2\theta = 1\).
Remember reciprocal identities: \(\csc\theta = 1/\sin\theta\), \(\sec\theta = 1/\cos\theta\), \(\cot\theta = 1/\tan\theta\).

Answer 1

The Correct Option is B

Let the expression be E. \[ E = \frac{\sin x}{1+\cos x} + \frac{1+\cos x}{\sin x} \] Combine the fractions with a common denominator \((1+\cos x)\sin x\): \[ E = \frac{\sin x \cdot \sin x + (1+\cos x)(1+\cos x)}{(1+\cos x)\sin x} \] \[ E = \frac{\sin^2 x + (1+\cos x)^2}{(1+\cos x)\sin x} \] Expand the numerator: \((1+\cos x)^2 = 1^2 + 2(1)(\cos x) + \cos^2 x = 1 + 2\cos x + \cos^2 x\). So, the numerator is \(\sin^2 x + 1 + 2\cos x + \cos^2 x\). Using the identity \(\sin^2 x + \cos^2 x = 1\): Numerator = \((\sin^2 x + \cos^2 x) + 1 + 2\cos x = 1 + 1 + 2\cos x = 2 + 2\cos x\). Factor out 2 from the numerator: Numerator = \(2(1+\cos x)\). Now substitute back into E: \[ E = \frac{2(1+\cos x)}{(1+\cos x)\sin x} \] Assuming \(1+\cos x \neq 0\) (i.e., \(\cos x \neq -1\), so \(x \neq (2k+1)\pi\)) and \(\sin x \neq 0\) (i.e., \(x \neq k\pi\)), we can cancel the term \((1+\cos x)\): \[ E = \frac{2}{\sin x} \] Since \(\csc x = \frac{1}{\sin x}\), \[ E = 2 \csc x \] This matches option (b). \[ \boxed{2 \csc x} \]