Question

In $△ABC,\frac{\sin 2A+\sin 2B+\sin 2C}{\sin A+\sin B+\sin C}$ is equal to :-

• A. $8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
• B. $8\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$
• C. $8\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}$
• D. $8\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$

Answer 1

The correct option is A $8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

$\frac{sin2A+sin2B+sin2C}{sinA+sinB+sinC}$

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

$A+B+C=\pi$ For triangle

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

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

$=\frac{2sinC[2sinA.sinB]}{2cos\frac{C}{2}[2cos\frac{A}{2}.cos\frac{B}{2}]}$

$=\frac{2\times 2sin\frac{C}{2}.cos\frac{C}{2}\times 2\times sin\frac{A}{2}.cos\frac{A}{2}\times 2\times 2sin\frac{B}{2}.cos\frac{B}{2}}{2cos\frac{C}{2}.2cos\frac{A}{2}.cos\frac{B}{2}}$

$=\frac{2^5}{2^2}\times sin\frac{A}{2}.sin\frac{B}{2}.sin\frac{C}{2}$

$=8sin\frac{A}{2}.sin\frac{B}{2}.sin\frac{C}{2}$