Question

If $A+B+C=\pi$ then the expression $\frac{\sin 2A+\sin 2B-\sin 2C}{\sin 2A+\sin 2B+\sin 2C}$ reduces to

Answer 1

Given that

$A+B+C=\pi$

Now

$\frac{\sin 2A+\sin 2B-sin2C}{\sin 2A+\sin 2B+sin2C}$

$\frac{2\sin \frac{2A+2B}{2}\cos \frac{2A-2B}{2}-\sin 2C}{2\sin \frac{2A+2B}{2}\cos \frac{2A-2B}{2}+\sin 2C}$ [using the formula of $sin2A+sin2B$]

$\frac{2\sin (A+B)\cos \left(A-B\right)-\sin 2C}{2\sin (A+B)\cos \left(A-B\right)+\sin 2C}$

$=\frac{2\sin C\cos \left(A-B\right)-2\sin C\cos C}{2\sin C\cos \left(A-B\right)+2\sin C\cos C}$ $\because A+B=\pi -C$

$=\frac{2\sin C\left[\cos \left(A-B\right)-\cos C\right]}{2\sin C\left[\cos \left(A-B\right)+\cos C\right]}$

$=\frac{\cos \left(A-B\right)+\cos \left(A+B\right)}{\cos \left(A-B\right)-\cos \left(A+B\right)}$

$=\frac{2\cos \frac{\left(A-B+A+B\right)}{2}\cos \frac{\left(A-B-A+B\right)}{2}}{2\sin \frac{\left(A-B+A+B\right)}{2}\sin \frac{\left(A-B-A+B\right)}{2}}$ [using the formula of $\cos \left(A-B\right)\pm \cos \left(A+B\right)$]

$=\frac{2\cos \left(\frac{2A}{2}\right)\cos \left(\frac{2B}{2}\right)}{2\sin \left(\frac{2A}{2}\right)sin\left(\frac{2B}{2}\right)}$

$=\frac{\cos A\cos B}{\sin A\sin B}$

$=\cot A\cot B$

Hence, it is the reduces from of the given equations.