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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]4 Marks
Question
In $\triangle A B C$, if $\angle A=\frac{\pi}{2}$, then prove that $\sin (B-C)=\frac{b^2-c^2}{b^2+c^2}$
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Answer

$ \text { In } \triangle ABC , \angle A =\frac{\pi}{2} \quad \ldots . . . \text { [Given] }$
$\therefore \sin A =\frac{\sin \pi}{2}=1$
$\text { and } A + B + C =\pi$
$\therefore B + C =\frac{\pi}{2}$
$\therefore B =\frac{\pi}{2}- C \text { and } C =\frac{\pi}{2}- B$
$\therefore \sin B =\sin \left(\frac{\pi}{2}- C \right)$
$=\cos C \text { and } \sin C$
$=\sin \left(\frac{\pi}{2}- B \right)$
$=\cos B \ldots \ldots . . \text { (i) } $
In $\triangle ABC$ by sine rule, we have
$ \frac{ a }{\sin A }=\frac{ b }{\sin B }=\frac{ c }{\sin C }$
$\therefore \frac{ a }{1}=\frac{ b }{\sin B }=\frac{ c }{\sin C }$
$\therefore b = a \sin B , c = a \sin C \quad \ldots \ldots . . \text { (ii) }$
$\therefore \text { R.H.S. }=\frac{ b ^2- c ^2}{ b ^2+ c ^2} $
$=\frac{a^2 \sin ^2 B-a^2 \sin ^2 C}{a^2 \sin ^2 B+a^2 \sin ^2 C} \ldots \ldots .[\text { From (ii)] }$
$=\frac{\sin ^2 B-\sin ^2 C}{\sin ^2 B+\sin ^2 C}$
$=\frac{\sin B \cdot(\sin B)-\sin C \cdot(\sin C)}{\sin ^2 B+\sin ^2 C}$
$=\frac{\sin B \cos C-\sin C \cos B}{\sin ^2 B+\cos ^2 B} \ldots \ldots . .[\text { From (i) ] }$
$=\frac{\sin (B-C)}{1}$
$=\sin (B-C)$
$=\text { L.H.S. }$

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