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If $\cot \left(\frac{A}{2}\right)=\sqrt{\frac{1+a}{1-a}} \cdot \cot \left(\frac{\theta}{2}\right)$, then $\cos \theta=$
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Verified Answer
The correct answer is:
$\frac{(\cos A)-a}{1-\operatorname{acos} A}$
$\begin{aligned} & \text { We have, } \cot \frac{A}{2}=\sqrt{\frac{1+a}{1-a}} \cot \frac{\theta}{2} \\ & \Rightarrow \quad \cot ^2 \frac{A}{2}=\left(\frac{1+a}{1-a}\right) \cot ^2 \frac{\theta}{2}\end{aligned}$
$\begin{aligned}
& \Rightarrow\left(\frac{\cos ^2 \frac{A}{2}}{\sin ^2 \frac{A}{2}}\right)\left(\frac{1-a}{1+a}\right)=\frac{\cos ^2 \theta / 2}{\sin ^2 \theta / 2} \\
& \Rightarrow \frac{(1+\cos A)(1-a)}{(1-\cos A)(1+a)}=\frac{1+\cos \theta}{1-\cos \theta} \\
& \Rightarrow \frac{1-a-a \cos A+\cos A}{1+a-a \cos A-\cos A}=\frac{1+\cos \theta}{1-\cos \theta}
\end{aligned}$
Apply componendo and dividendo,
$\Rightarrow \frac{2(1-a \cos A)}{2(\cos A-a)}=\frac{2}{2 \cos \theta} \Rightarrow \cos \theta=\frac{\cos A-a}{1-a \cos A}$
$\begin{aligned}
& \Rightarrow\left(\frac{\cos ^2 \frac{A}{2}}{\sin ^2 \frac{A}{2}}\right)\left(\frac{1-a}{1+a}\right)=\frac{\cos ^2 \theta / 2}{\sin ^2 \theta / 2} \\
& \Rightarrow \frac{(1+\cos A)(1-a)}{(1-\cos A)(1+a)}=\frac{1+\cos \theta}{1-\cos \theta} \\
& \Rightarrow \frac{1-a-a \cos A+\cos A}{1+a-a \cos A-\cos A}=\frac{1+\cos \theta}{1-\cos \theta}
\end{aligned}$
Apply componendo and dividendo,
$\Rightarrow \frac{2(1-a \cos A)}{2(\cos A-a)}=\frac{2}{2 \cos \theta} \Rightarrow \cos \theta=\frac{\cos A-a}{1-a \cos A}$
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