Q. If $\cos{\theta = \cos{\alpha\cos\beta}}$, then $\tan\left( \frac{\theta + \alpha}{2} \right)\tan\left( \frac{\theta - \alpha}{2} \right)$ is equal to

A) $\tan^{2}\frac{\alpha}{2}$
B) $\tan^{2}\frac{\beta}{2}$
C) $\tan^{2}\frac{\theta}{2}$
D) $\cot^{2}\frac{\beta}{2}$
Explanation

(b)
$\tan{\left( \frac{\theta + \alpha}{2} \right).\tan\left( \frac{\theta - \alpha}{2} \right)}$
$= \frac{\tan^{2}{\frac{\theta}{2} - \tan^{2}\frac{\alpha}{2}}}{1 - \tan^{2}{\frac{\theta}{2}\tan^{2}\frac{\alpha}{2}}}$
$= \frac{\sin^{2}{\frac{\theta}{2}\cos^{2}{\frac{\alpha}{2} - \sin^{2}{\frac{\alpha}{2}\cos^{2}\frac{\theta}{2}}}}}{\cos^{2}{\frac{\alpha}{2}\cos^{2}{\frac{\theta}{2} - \sin^{2}{\frac{\alpha}{2}\sin^{2}\frac{\theta}{2}}}}}$
$= \frac{\cos^{2}{\frac{\alpha}{2} - \cos^{2}{\frac{\theta}{2}\cos^{2}{\frac{\alpha}{2} - \cos^{2}{\frac{\theta}{2} + \cos^{2}{\frac{\alpha}{2}\cos^{2}\frac{\theta}{2}}}}}}}{\cos^{2}{\frac{\theta}{2} - \sin^{2}{\frac{\alpha}{2}\cos^{2}{\frac{\theta}{2} - \sin^{2}{\frac{\alpha}{2} + \sin^{2}{\frac{\alpha}{2}\cos^{2}\frac{\theta}{2}}}}}}}$
$= \frac{\cos^{2}{\frac{\alpha}{2} - \cos^{2}\frac{\theta}{2}}}{\cos^{2}{\frac{\theta}{2} - \sin^{2}\frac{\alpha}{2}}} = \frac{(\cos{\alpha - \cos\theta})}{(\cos{\alpha + \cos\theta})}$
$= \frac{\cos{\alpha(1 - \cos\beta)}}{\cos{\alpha(1 + \cos\beta)}} = \tan^{2}\frac{\beta}{2}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.