Question

If $A+B+C=180°$, then the value of $(cotB+cotC)(cotC+cotA)(cotA+cotB)$will be

• A. $secAsecBsecC$
• B. $\cos ecA\cos ecB\cos ecC$
• C. $\tan A\tan B\tan C$
• D. $1$

Answer 1

The correct option is B

$\cos ecA\cos ecB\cos ecC$

Explanation for correct option.

Given, $A+B+C=180°$

Now,

$\begin{array}{rcl}A+B+C& =& 180°......\left(i\right)\\ & & (cotB+cotC)(cotC+cotA)(cotA+cotB)\\ & =& \left[\right(\frac{\cos B}{\sin B})+(\frac{\cos C}{\sin C}\left)\right]\left[\right(\frac{\cos C}{\sin C})+(\frac{\cos A}{\sin A}\left)\right]\left[\right(\frac{\cos A}{\sin A})+(\frac{\cos B}{\sin B}\left)\right]\\ & =& \left[\frac{(\cos B\sin C+\cos C\sin B}{\sin B\sin C}\right)\left]\right[\frac{(\sin A\cos C+\cos A\sin C)}{\sin A\sin C}\left]\right[\left(\frac{\cos A\sin B+\sin A\cos B}{\sin A\sin B}\right)]\\ & =& \left[\frac{\sin (B+C)}{\sin B\sin C}\right]\left[\frac{\sin (A+C)}{\sin A\sin C}\right]\left[\frac{\sin (A+B)}{\sin A\sin B}\right]\\ & =& \left[\frac{\sin (180°–A)}{\sin B\sin C}\right]\left[\frac{\sin (180°-B)}{\sin A\sin C}\right]\left[\frac{\sin (180°–C)}{\sin A\sin B}\right].....from\left(i\right)\\ & =& \left[\frac{\sin A}{\sin B\sin C}\right]\left[\frac{\sin B}{\sin A\sin C}\right]\left[\frac{\sin C}{\sin A\sin B}\right]\\ & =& \frac{1}{\sin A\sin B\sin C}\\ & =& \cos ecA\cos ecB\cos ecC\\ & & \\ & & \end{array}$

Hence, the correct option is $\left(B\right)$