^2 ~A+ ^2 ~B- ^2 C=2 A B C

Question

$\sin ^2 \mathrm{~A}+\sin ^2 \mathrm{~B}-\sin ^2 \mathrm{C}=2 \sin \mathrm{A} \sin \mathrm{B} \cos \mathrm{C}$

✓

Answer

We know that, $\sin ^2=\frac{1-\cos 2 \theta}{2}$
L.H.S.
$
\begin{aligned}
& =\sin ^2+\sin ^2 B+\sin ^2 C \\
& =\frac{1-\cos 2 A}{2}+\frac{1-\cos 2 B}{2}-\sin ^2 C \\
& =\frac{1}{2}[2-(\cos 2 A+\cos 2 B)]-\sin ^2 C \\
& =\frac{1}{2}\left[2-2 \cdot \cos \left(\frac{2 A+2 B}{2}\right) \cdot \cos \left(\frac{2 A-2 B}{2}\right)\right]
\end{aligned}$
$\begin{aligned}
& -\sin ^2 C \\
& =1-\cos (A+B) \cdot \cos (A-B)-\sin ^2 C \\
& =\left(1-\sin ^2 C\right)-\cos (A+B) \cdot \cos (A-B) \\
& =\cos ^2 C-\cos (A+B) \cdot \cos (A-B) \\
& \therefore \cos (A+B)=\cos (\text { it }-C) \\
& \therefore \cos (A+B)=-\cos C \ldots \text { (i) } \\
& \therefore \text { L.H.S. }=\cos ^2 \mathrm{C}+\cos \mathrm{C} \cdot \cos (\mathrm{A}-\mathrm{B}) \\
& \text {... [From (i)] } \\
& =\cos C[\cos C+\cos (A-B)] \\
& =\cos C[-\cos (A+B)+\cos (A-B)] \\
& \text {... [From (i)] } \\
& =\cos C[\cos (A-B)-\cos (A+B)] \\
& =\cos C(2 \sin A \cdot \sin B) \\
& =2 \sin A \cdot \sin B \cdot \cos C \\
& =\mathrm{R} . \mathrm{H} . \mathrm{S} \text {. } \\
\end{aligned}$
[Note: The question has been modified.]

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