If \( \tan\frac{x}{2}=\sqrt{\frac{1-e}{1+e}}\tan\frac{\alpha}{2} \), Find \( \cos\alpha \)
Question
If
\[ \tan\frac{x}{2} = \sqrt{\frac{1-e}{1+e}} \tan\frac{\alpha}{2}, \]
then \(\cos\alpha\) is equal to
(a) \(\dfrac{1-e\cos x}{\cos x+e}\)
(b) \(\dfrac{1+e\cos x}{\cos x-e}\)
(c) \(\dfrac{1-e\cos x}{\cos x-e}\)
(d) \(\dfrac{\cos x-e}{1-e\cos x}\)
Solution
Let
\[ t=\tan\frac{\alpha}{2} \]
Then
\[ \tan\frac{x}{2} = \sqrt{\frac{1-e}{1+e}}\,t \]
Squaring,
\[ \tan^2\frac{x}{2} = \frac{1-e}{1+e}t^2 \]
Using
\[ \tan^2\frac{\theta}{2} = \frac{1-\cos\theta}{1+\cos\theta}, \]
we get
\[ \frac{1-\cos x}{1+\cos x} = \frac{1-e}{1+e} \cdot \frac{1-\cos\alpha}{1+\cos\alpha} \]
Cross-multiplying,
\[ (1+\cos\alpha)(1-e)(1+\cos x) = (1-\cos\alpha)(1+e)(1-\cos x) \]
On simplifying and collecting terms containing \(\cos\alpha\),
\[ \cos\alpha = \frac{\cos x-e} {1-e\cos x} \]
Final Answer
\[ \boxed{\cos\alpha=\frac{\cos x-e}{1-e\cos x}} \]
Hence, the correct option is (d).