Question

If $A+B+C={180}^o$, then the value of $(\cot B+\cot C)(\cot C+\cot A)(\cot A+\cot B)$ will be

• A. $\sec A\sec B\sec C$
• B. $\csc A\csc B\csc C$
• C. $\tan A\tan B\tan C$
• D. $1$

Answer 1

The correct option is B $\csc A\csc B\csc C$

Given : $A+B+C={180}^o$ ........ (i)

Let us solve $(\cot B+\cot C)(\cot C+\cot A)(\cot A+\cot B)$

$=(\frac{\cos B}{\sin B}+\frac{\cos C}{\sin C})(\frac{\cos C}{\sin C}+\frac{\cos A}{\sin A})(\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B})$

$=\left(\frac{\cos B\sin C+\sin B\cos C}{\sin B\sin C}\right)\left(\frac{\cos C\sin A+\sin C\cos A}{\sin C\sin A}\right)\left(\frac{\cos A\sin B+\sin A\cos B}{\sin A\sin B}\right)$

$=\left(\frac{\sin (B+C)}{\sin B\sin C}\right)\left(\frac{\sin (A+C)}{\sin C\sin A}\right)\left(\frac{\sin (A+B)}{\sin A\sin B}\right)$

$=\left(\frac{\sin ({180}^o-A)}{\sin B\sin C}\right)\left(\frac{\sin ({180}^o-B)}{\sin C\sin A}\right)\left(\frac{\sin ({180}^o-C)}{\sin A\sin B}\right)$ ......... [Using (i)]

$=\left(\frac{\sin A}{\sin B\sin C}\right)\left(\frac{\sin B}{\sin C\sin A}\right)\left(\frac{\sin C}{\sin A\sin B}\right)$ ........ $[\because \sin ({180}^o-x)=\sin x]$

$=\frac{1}{\sin A\sin B\sin C}$

$=\csc A\csc B\csc C$

Hence, $(\cot B+\cot C)(\cot C+\cot A)(\cot A+\cot B)=\csc A\csc B\csc C$