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. $cosecAcosecBcosecC$
• C. $\tan A\tan B\tan C$
• D. $1$

Answer 1

Answer: (B)

$cosecAcosecBcosecC$

$A+B+C={180}^o$ [ Given ]

$\Rightarrow$ $B+C={180}^o-A$ ---- ( 1 )

$\Rightarrow$ $A+C={180}^o-B$ ---- ( 2 )

$\Rightarrow$ $A+B={180}^o-C$ ---- ( 3 )

$(\cot B+\cot C)(\cot C+\cot A)(\cot A+\cot B)$

$\Rightarrow$ $[\frac{\cos B}{\sin B}+\frac{\cos C}{\sin C}][\frac{\cos C}{\sin C}+\frac{\cos A}{\sin A}][\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}]$

$\Rightarrow$ $\left[\frac{\cos B\sin C+\cos C\sin B}{\sin B\sin C}\right]\left[\frac{\cos C\sin A+\sin C\cos A}{\sin A\sin C}\right]\left[\frac{\sin B\cos A+\cos B\sin A}{\sin A\sin B}\right]$

We know, $sin(x+y)=\sin x\cos y+\cos x\sin y$

$\Rightarrow$ $\left[\frac{\sin (B+C)}{\sin B\sin C}\right]\left[\frac{\sin (A+C)}{\sin A\sin C}\right]\left[\frac{\sin (A+B)}{\sin A\sin B}\right]$

From ( 1 ), ( 2 ) and ( 3 )

$\Rightarrow$ $\left[\frac{\sin ({180}^o-A)}{\sin B\sin C}\right]\left[\frac{\sin ({180}^o-B)}{\sin A\sin C}\right]\left[\frac{\sin ({180}^o-C)}{\sin A\sin B}\right]$

We know, $\sin ({180}^o-\theta )=\sin \theta$

$\Rightarrow$ $\frac{\sin A}{\sin B\times \sin C}\times \frac{\sin B}{\sin A\times \sin C}\times \frac{\sin C}{\sin A\times \sin B}$

$\Rightarrow$ $\frac{1}{\sin A}.\frac{1}{\sin B}.\frac{1}{\sin C}$

$\Rightarrow$ $cosecAcosecBcosecC$