AP EAMCET · Maths · Trigonometric Ratios & Identities
In \(\triangle \mathrm{ABC}, \frac{\sin 2 \mathrm{~A}+\sin 2 \mathrm{~B}+\sin 2 \mathrm{C}}{\cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}-1}=\)
- A \(2[\sin A+\sin B+\sin C]\)
- B \(\sin A+\sin B+\sin C\)
- C \(4[\sin A+\sin B+\sin C]\)
- D \(8[\sin A+\sin B+\sin C]\)
Answer & Solution
Correct Answer
(A) \(2[\sin A+\sin B+\sin C]\)
Step-by-step Solution
Detailed explanation
\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}…
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