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The value of \(\frac{\sin\theta}{1+\cos\theta} + \frac{1+\cos\theta}{\sin\theta}\) is:

  • A.\(2\tan\theta\)
  • B.\(2\sec\theta\)
  • C.\(2\csc\theta\)
  • D.\(2\cot\theta\)

Answer: C

We find a common denominator, which is \(\sin\theta(1+\cos\theta)\).

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

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

Using \(\sin^2\theta + \cos^2\theta = 1\), the numerator becomes \(1 + 1 + 2\cos\theta = 2 + 2\cos\theta = 2(1+\cos\theta)\).

= \(\frac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)} = \frac{2}{\sin\theta} = 2\csc\theta\).

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