Question

If $\frac{tan(\alpha +\beta -\gamma )}{tan(\alpha -\beta +\gamma )}=\frac{tan\gamma }{tan\beta }$ , then either $sin(\beta -\gamma )=0$ or $sin2\alpha +sin2\beta +sin2\gamma =0$

• A. True
• B. False

Answer 1

The correct option is A True

Let $B-C=X$

so, $\tan (A+X)/\tan (A-X)=\tan C/\tan B$

$\to \sin (A+X)\cos (A-X)/\cos (A+X)\sin (A-X)=\frac{\sin C\cos B}{\cos C\sin B}$

adjust a factor $2$ in both sides with Numerator and Denominator.

$\to \sin 2A+\sin 2X/\sin 2A-\sin 2X=\frac{-\sin (C+B)+\sin (C+B)}{\sin (B+C)+\sin (B-C)}$

apply componendo- dividendo -

$\frac{\sin 2A}{\sin 2X}=\frac{\sin (B+C)}{-\sin (B-C)}\Rightarrow \frac{\sin 2A}{2\sin x\cos x}=\frac{\sin (B+C)}{\sin x}$

$\Rightarrow \sin 2A=-2\sin (B+C)\cos (B-C)=-[\sin 2B+\sin 2C]$.