Search any question & find its solution
Question:
Answered & Verified by Expert
In $\triangle \mathrm{ABC}, \frac{\sin 2 \mathrm{~A}+\sin 2 \mathrm{~B}+\sin 2 \mathrm{C}}{\cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}-1}=$
Options:
Solution:
2461 Upvotes
Verified Answer
The correct answer is:
$2[\sin A+\sin B+\sin C]$
$\begin{aligned}
& \text {} \frac{\sin 2 A+\sin 2 B+\sin 2 C}{\cos A+\cos B+\cos C-1} \\
& =\frac{2 \sin (\mathrm{A}+B) \cos (\mathrm{A}-\mathrm{B})+2 \sin \mathrm{Ccos} C}{2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)+\left(1-2 \sin ^2 \frac{C}{2}\right)-1} \\
& =\frac{2 \sin C(\cos (\mathrm{A}-\mathrm{B})+\cos C)}{2 \sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)\right]} \\
& =\frac{\sin C(\cos (\mathrm{A}-\mathrm{B})+\cos (A+B))}{2 \sin \frac{C}{2} \sin \frac{B}{2} \sin \frac{A}{2}} \\
& =\frac{2 \sin C \sin B \sin A}{2 \sin \frac{C}{2} \cos \frac{B}{2} \sin \frac{A}{2}} \\
& =8 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}...(i)
\end{aligned}$
Now, $\sin A+\sin B+\sin C$
$\begin{aligned}
& =2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)+\sin C \\
& =2 \cos \left(\frac{C}{2}\right)\left[\cos \left(\frac{A-B}{2}\right)+\sin \left(\frac{\pi}{2}-\left(\frac{A+B}{2}\right)\right)\right] \\
& =2 \cos \left(\frac{C}{2}\right)\left[\cos \left(\frac{A-B}{2}\right)+\cos \left(\frac{A+B}{2}\right)\right] \\
& =4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}...(ii)
\end{aligned}$
From equations (i) \& (ii), we get
$\frac{\sin 2 A+\sin 2 B+\sin 2 C}{\cos A+\cos B+\cos C}=2[\sin A+\sin B+\sin C]$
& \text {} \frac{\sin 2 A+\sin 2 B+\sin 2 C}{\cos A+\cos B+\cos C-1} \\
& =\frac{2 \sin (\mathrm{A}+B) \cos (\mathrm{A}-\mathrm{B})+2 \sin \mathrm{Ccos} C}{2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)+\left(1-2 \sin ^2 \frac{C}{2}\right)-1} \\
& =\frac{2 \sin C(\cos (\mathrm{A}-\mathrm{B})+\cos C)}{2 \sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)\right]} \\
& =\frac{\sin C(\cos (\mathrm{A}-\mathrm{B})+\cos (A+B))}{2 \sin \frac{C}{2} \sin \frac{B}{2} \sin \frac{A}{2}} \\
& =\frac{2 \sin C \sin B \sin A}{2 \sin \frac{C}{2} \cos \frac{B}{2} \sin \frac{A}{2}} \\
& =8 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}...(i)
\end{aligned}$
Now, $\sin A+\sin B+\sin C$
$\begin{aligned}
& =2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)+\sin C \\
& =2 \cos \left(\frac{C}{2}\right)\left[\cos \left(\frac{A-B}{2}\right)+\sin \left(\frac{\pi}{2}-\left(\frac{A+B}{2}\right)\right)\right] \\
& =2 \cos \left(\frac{C}{2}\right)\left[\cos \left(\frac{A-B}{2}\right)+\cos \left(\frac{A+B}{2}\right)\right] \\
& =4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}...(ii)
\end{aligned}$
From equations (i) \& (ii), we get
$\frac{\sin 2 A+\sin 2 B+\sin 2 C}{\cos A+\cos B+\cos C}=2[\sin A+\sin B+\sin C]$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.