Trigonometric Functions — Maths STD 12 Science — Question

$\therefore c^2=a^2+b^2 \ldots(1)$

By sine rule,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

$\therefore \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin 90^{\circ}}$

$\therefore \frac{a}{\sin A}=\frac{b}{\sin B}=c$

$\cdots\left[\because \sin 90^{\circ}=1\right]$

$\therefore \sin \mathrm{A}=\frac{a}{c}$ and $\sin \mathrm{B}=\frac{b}{c}$

$\ldots(2)$

$\begin{aligned} \text { LHS } & =\sin (A-B) \\ & =\sin A \cos B-\cos A \sin B\end{aligned}$

$=\frac{a}{c} \cos B-\frac{b}{c} \cos A$

$\ldots$ [By (2)]

$=\frac{a}{c}\left(\frac{c^2+a^2-b^2}{2 c a}\right)-\frac{b}{c}\left(\frac{b^2+c^2-a^2}{2 b c}\right)$

$\begin{aligned} & =\frac{c^2+a^2-b^2}{2 c^2}-\frac{b^2+c^2-a^2}{2 c^2} \\ & =\frac{c^2+a^2-b^2-b^2-c^2+a^2}{2 c^2}\end{aligned}$

$\begin{aligned} & =\frac{2 a^2-2 b^2}{2 c^2}=\frac{a^2-b^2}{c^2} \\ & =\frac{a^2-b^2}{a^2+b^2} \\ & =\text { RHS. }\end{aligned}$

$\ldots[$ By (1)]