Question

The value of \(\dfrac{\cos\theta}{1+\sin\theta}\) is equal to

• A. \(\tan\left(\frac{\theta}{2}-\frac{\pi}{4}\right)\)
• B. \(\tan\left(-\frac{\pi}{4}-\frac{\theta}{2}\right)\)
• C. \(\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\)
• D. \(\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\)

Hint:

\(\dfrac{\cos\theta}{1+\sin\theta}=\dfrac{1-\sin\theta}{\cos\theta}=\tan\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right)\). Multiply by \(1-\sin\theta\) to simplify.

Answer 1

The Correct Option is C

Step 1: Start with expression.
\[ \frac{\cos\theta}{1+\sin\theta} \] Step 2: Multiply numerator and denominator by \((1-\sin\theta)\).
\[ \frac{\cos\theta}{1+\sin\theta}\cdot\frac{1-\sin\theta}{1-\sin\theta} = \frac{\cos\theta(1-\sin\theta)}{1-\sin^2\theta} \] Step 3: Simplify denominator.
\[ 1-\sin^2\theta=\cos^2\theta \] So:
\[ \frac{\cos\theta(1-\sin\theta)}{\cos^2\theta} = \frac{1-\sin\theta}{\cos\theta} \] Step 4: Recognize identity.
\[ \frac{1-\sin\theta}{\cos\theta} = \tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right) \] This is a standard trigonometric identity.
Final Answer: \[ \boxed{\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)} \]