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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II3 Marks
Question
In $\triangle A B C$ prove that
$\sin ^2 \mathrm{~A}+\sin ^2 \mathrm{~B}-\sin ^2 \mathrm{C}=2 \sin \mathrm{A} \sin \mathrm{B} \sin \mathrm{C} $
✓

Answer

$\begin{aligned} & \text { L.H.S. }=\sin ^2 \mathrm{~A}+\sin ^2 \mathrm{~B}-\sin ^2 \mathrm{C} \\ & =\frac{1-\cos 2 A}{2}+\frac{1-\cos 2 B}{2}-\sin ^2 C \\ & =\frac{1}{2}[2-\cos 2 \mathrm{~A}-\cos 2 \mathrm{~B}]-\sin ^2 \mathrm{C} \\ & =1-\frac{1}{2}[\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}]-\sin ^2 \mathrm{C} \\ & =1-\frac{1}{2} \cdot 2 \cos \left(\frac{2 A+2 B}{2}\right) \cos \left(\frac{2 A-2 B}{2}\right)-\sin ^2 C \\ & =1-\sin ^2 C-\cos (A+B)+\cos (A-B) \\ & =\cos { }^2 C-\cos [\pi-C] \cos (A-B) \\ & =\cos { }^2 C+\cos C \cos (A-B) \\ & =\cos C[\cos C+\cos (A-B)] \\ & =\cos C[\cos [\pi-(A+B)]+\cos (A-B)] \\ & =\cos C[-\cos (A+B)+\cos (A-B)] \\ & =\cos C[\cos (A-B)-\cos (A+B)]\end{aligned}$
$\begin{aligned} & =\cos C \cdot 2 \sin \left(\frac{A-B+A+B}{2}\right) \sin \left(\frac{A+B-A+B}{2}\right) \\ & =2 \cos C \sin A \sin B \\ & =2 \sin A \sin B \cos C \\ & =\text { R. H.S. }\end{aligned}$

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