Question : If $\tan \frac{A}{2}=x$, then find $x$.
Option 1: $\frac{\sqrt{1+\cos A}}{\sqrt{1-\cos A}}$
Option 2: $\frac{\sqrt{1-\sin A}}{\sqrt{1+\cos A}}$
Option 3: $\frac{\sqrt{1-\cos A}}{\sqrt{1+\cos A}}$
Option 4: $\frac{\sqrt{\cos A-1}}{\sqrt{1+\cos A}}$
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Correct Answer: $\frac{\sqrt{1-\cos A}}{\sqrt{1+\cos A}}$
Solution :
Given: $\tan\frac{A}{2}=x$
We know,
$\cos A=1-2\sin^2(\frac{A}{2})$
⇒ $\sin(\frac{A}{2})=\sqrt{\frac{1–\cos A}{2}}$
Also, $\cos A=2\cos^2(\frac{A}{2})–1$
⇒ $\cos(\frac{A}{2})=\sqrt{\frac{1+\cos A}{2}}$
Now, $\tan\frac{A}{2}=\frac{\sin(\frac{A}{2})}{\cos(\frac{A}{2})}$
⇒ $x=\sqrt{\frac{1–\cos A}{2}}×\sqrt{\frac{2}{1+\cos A}}$
$\therefore x=\frac{\sqrt{1-\cos A}}{\sqrt{1+\cos A}}$
Hence, the correct answer is $\frac{\sqrt{1-\cos A}}{\sqrt{1+\cos A}}$.
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