If $\cos \theta=\frac{\cos \alpha-\cos \beta}{1-\cos \alpha \cos \beta}$, then one of the values of $\tan…
If $\cos \theta=\frac{\cos \alpha-\cos \beta}{1-\cos \alpha \cos \beta}$, then one of the values of $\tan \frac{\theta}{2}$ is
- $\cot \frac{\beta}{2} \tan \frac{\alpha}{2}$
- $\tan \alpha \tan \frac{\beta}{2}$
- $\tan \frac{\beta}{2} \cot \frac{\alpha}{2}$
- $\tan ^2 \frac{\alpha}{2} \tan ^2 \frac{\beta}{2}$
Solution
We have, $\cos \theta=\frac{\cos \alpha-\cos \beta}{1-\cos \alpha \cos \beta}$
Appling componendo and dividendo, we get
$\begin{aligned}
& \frac{\cos \theta+1}{\cos \theta-1}=\frac{\cos \alpha-\cos \beta+1-\cos \alpha \cos \beta}{\cos \alpha-\cos \beta-1+\cos \alpha \cos \beta} \\
& \frac{\cos \theta+1}{\cos \theta-1}=\frac{(\cos \alpha+1)(1-\cos \beta)}{(\cos \alpha-1)(\cos \beta+1)} \\
& \frac{1+\cos \theta}{1-\cos \theta}=\frac{(1+\cos \alpha)(1-\cos \beta)}{(1-\cos \alpha)(1+\cos \beta)}
\end{aligned}$
$\begin{aligned} & \frac{2 \cos ^2 \frac{\theta}{2}}{2 \sin ^2 \frac{\theta}{2}}=\frac{2 \cos ^2 \frac{\alpha}{2}}{2 \sin ^2 \frac{\alpha}{2}} \frac{2 \sin ^2 \frac{\beta}{2}}{2 \cos ^2 \frac{\beta}{2}} \\ & \cot ^2 \frac{\theta}{2}=\cot ^2 \frac{\alpha}{2} \tan ^2 \frac{\beta}{2} \\ & \tan ^2 \frac{\theta}{2}=\tan ^2 \frac{\alpha}{2} \cot ^2 \frac{\beta}{2} \\ & \tan \frac{\theta}{2}=\tan \frac{\alpha}{2} \cot \frac{\beta}{2}\end{aligned}$
Asked in: AP EAMCET 2016
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