Question

In $△ABC$, if $\cot x=\cot A+\cot B+\cot C$, then $\sin (A-x)\sin (B-x)\sin (C-x)$ is

• A. ${\sin }^{3}x$
• B. ${\cos }^{3}x$
• C. ${\sin }^{2}x$
• D. $3\sin x$

Answer 1

The correct option is A ${\sin }^{3}x$

Let
$\cot x=\cot A+\cot B+\cot C$
$\Rightarrow \cot x-\cot A=\cot B+\cot C\Rightarrow \frac{\sin (A-x)}{\sin A\sin x}=\frac{\sin (B+C)}{\sin B\sin C}\Rightarrow \sin (A-x)=\frac{{\sin }^{2}A\sin x}{\sin B\sin C}\cdots \left(1\right)$
Similarly,
$\sin (B-x)=\frac{{\sin }^{2}B\sin x}{\sin A\sin C}\cdots \left(2\right)$
$\sin (C-x)=\frac{{\sin }^{2}C\sin x}{\sin A\sin B}\cdots \left(3\right)$

Multiplying equation $\left(1\right),\left(2\right)$ and $\left(3\right)$,
$\sin (A-x)\sin (B-x)\sin (C-x)={\sin }^{3}x$