Question

If $\cos (\alpha +\beta )\sin (\gamma +\theta )=\cos (\alpha -\beta )\sin (\gamma -\theta )$ then $\tan \alpha \tan \beta \tan \gamma$=

• A. $tan\theta$
• B. $ta{n}^2\theta$
• C. $arctan\theta$
• D. Independent of $\theta$

Answer 1

Answer: (A)

$tan\theta$

Given

$\cos (\alpha +\beta )\sin (\gamma +\theta )=\cos (\alpha -\beta )\sin (\gamma -\theta )$

$\frac{\cos (\alpha +\beta )}{\cos (\alpha -\beta )}=\frac{\sin (\gamma -\theta )}{\sin (\gamma +\theta )}$

Applying Compoundo and Dividendo Rule.

$\frac{\cos (\alpha +\beta )+\cos (\alpha -\beta )}{\cos (\alpha +\beta )-\cos (\alpha -\beta )}=\frac{\sin (\gamma -\theta )+\sin (\gamma +\theta )}{\sin (\gamma -\theta )-\sin (\gamma +\theta )}$

$⟹\frac{2\cos \alpha \cos \beta }{-2\sin \alpha \sin \beta }=\frac{2\sin \gamma \cos \theta }{-2\cos \gamma \sin \theta }$

$\cot \alpha \cot \beta =\tan \gamma \cot \theta$

$⟹\tan \alpha \tan \beta \tan \gamma =\tan \theta$