With usual notations, In $\triangle \mathrm{ABC}, \angle \mathrm{C}=90^{\circ}$, then the value of $\sin…
With usual notations, In $\triangle \mathrm{ABC}, \angle \mathrm{C}=90^{\circ}$, then the value of $\sin (\mathrm{A}-\mathrm{B})$ is
- $\frac{a^2+b^2}{a^2-b^2}$
- $\frac{a^2-b^2}{a^2+b^2}$
- $\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{a}^2}$
- $\frac{a^2-b^2}{b^2}$
Solution
$\begin{aligned} & \because \angle C=90^{\circ} \\ & \Rightarrow \angle A+\angle B=90^{\circ} \\ & \Rightarrow \sin c=1 \text { and } \sin (A+B)=1\end{aligned}$
Also $\mathrm{c}^2=\mathrm{a}^2+\mathrm{b}^2$
$\begin{aligned} & \text { Now } \sin (A-B)=\frac{\sin (A-B) \cdot \sin (A+B)}{\sin ^2 C}=\frac{\sin ^2 A-\sin ^2 B}{\sin ^2 C} \\ & =\frac{k^2 a^2-k^2 b^2}{k^2 c^2}=\frac{a^2-b^2}{c^2}=\frac{a^2-b^2}{a^2+b^2}\end{aligned}$
Asked in: MHT CET 2022 (05 Aug Shift 2)
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