Question

Consider a triangle $ABC$ such that $\cot A+\cot B+\cot C=\cot \theta$, then $\sin (A-\theta )\sin (B-\theta )\sin (C-\theta )$ is,

• A. ${\tan }^3\theta$
• B. ${\cot }^3\theta$
• C. ${\sin }^3\theta$
• D. ${\cos }^3\theta$

Answer 1

The correct option is C ${\sin }^3\theta$

According to question:........................

$cot\theta =cotA+cotB+\cot C$

$cot\theta -cotA=cotB+\cot C$

$weknowthat:$

$\Rightarrow \frac{\cos \theta }{\sin \theta }-\frac{cosA}{sinA}=\frac{cosB}{\sin B}+\frac{cosC}{sinC}$

$\Rightarrow \frac{sinA\cos \theta -\sin \theta cosA}{\sin \theta sinA}=\frac{sinCcosB+\sin BcosC}{\sin BsinC}$

$\Rightarrow \frac{\sin (A-\theta )}{\sin \theta sinA}=\frac{\sin (C+B)}{\sin B\sin C}---------\left(1\right)$

$|Byusingformula:sin\left(A+B\right)=sinAcosB+cosAsinB$

$Nowweknowthat:ABCistriangle,$

$\Rightarrow A+B+C={180}^0$

$\Rightarrow B+C={180}^0-A$

$\therefore \sin (B+C)=\sin \left({180}^0-A\right)$

$\sin (B+C)=\sin A$

$Nowsubstituteinequ\left(i\right).....$

$\Rightarrow \sin (A-\theta )=\frac{\sin A}{\sin B\sin C}\times \sin \theta sinA$

$\sin (A-\theta )=\frac{{\sin }^{2}A\sin \theta }{\sin B\sin C}----------\left(2\right)$

$similerlywefindas................$

$\Rightarrow sin(B-\theta )=\frac{{\sin }^{2}B\sin \theta }{\sin B\sin C}--------\left(3\right)$

$\Rightarrow sin(C-\theta )=\frac{{\sin }^{2}C\sin \theta }{\sin B\sin C}--------\left(4\right)$

$nowmultiplyingequ\left(2\right),\left(3\right)\&\left(3\right)\Rightarrow sin(A-\theta )sin(B-\theta )sin(C-\theta )=\frac{{\sin }^{2}A\sin \theta }{\sin B\sin C}\times \frac{{\sin }^{2}B\sin \theta }{\sin B\sin C}\times \frac{{\sin }^{2}C\sin \theta }{\sin B\sin C}$

$={\sin }^3\theta$