Question

If $A + B + C = {180^ \circ }$ then the value of $\left( {\cot A + \cot B} \right)\left( {\cot B + \cot C} \right)\left( {\cot C + \cot A} \right)$ will be:
$\left( a \right){\text{ secA secB secC}}$
$\left( b \right){\text{ cosecA cosecB cosecC}}$
$\left( c \right){\text{ tanA tanB tanC}}$
$\left( d \right){\text{ 1}}$

Answer 1

As we know $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ and by using the identities and expanding the identities given in the question, we will be able to find the value of it. While solving, we will use the identities given in the question which is $A + B + C = {180^ \circ }$ .

Formula used:
Trigonometric formulas used in it are as
$\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
$\sin \left( {\theta + \phi } \right) = \sin \theta \cos \phi + \cos \theta \sin \phi$
$\dfrac{1}{{\sin \theta }} = \cos ec \theta$
Here,
$\theta \& \phi$, will be the angle

Complete step by step solution:
So we have the trigonometric identities given as $\left( {\cot A + \cot B} \right)\left( {\cot B + \cot C} \right)\left( {\cot C + \cot A} \right)$
So it can also be written as
$\Rightarrow \left( {\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\cos B}}{{\sin B}}} \right)\left( {\dfrac{{\cos B}}{{\sin B}} + \dfrac{{\cos C}}{{\sin C}}} \right)\left( {\dfrac{{\cos C}}{{\sin C}} + \dfrac{{\cos A}}{{\sin A}}} \right)$
Now by taking the LCM of the term which are in braces individually, we get
$\Rightarrow \left( {\dfrac{{\cos A\sin B + \cos B\sin A}}{{\sin A\sin B}}} \right)\left( {\dfrac{{\cos B\sin A + \cos C\sin B}}{{\sin B\sin C}}} \right)\left( {\dfrac{{\cos C\sin A + \cos A\sin C}}{{\sin C\sin A}}} \right)$
As we know that $\sin \left( {\theta + \phi } \right) = \sin \theta \cos \phi + \cos \theta \sin \phi$ , so the above equation will be written as
$\Rightarrow \left( {\dfrac{{\sin (A + B)}}{{\sin A\sin B}}} \right)\left( {\dfrac{{\sin (B + C)}}{{\sin B\sin C}}} \right)\left( {\dfrac{{\sin (C + A)}}{{\sin C\sin A}}} \right)$ , we will name it equation $1$
As it is given in the question that $A + B + C = {180^ \circ }$
So it can be written as
$\Rightarrow A + B = {180^ \circ } - C,{\text{ }}B + C = {180^ \circ } - A,{\text{ }}A + C = {180^ \circ } - B$
Now on substituting these values in the equation $1$ , we get
$\Rightarrow \left( {\dfrac{{\sin ({{180}^ \circ } - C)}}{{\sin A\sin B}}} \right)\left( {\dfrac{{\sin ({{180}^ \circ } - A)}}{{\sin B\sin C}}} \right)\left( {\dfrac{{\sin ({{180}^ \circ } - B)}}{{\sin C\sin A}}} \right)$
So on using the identities, we get
$\Rightarrow \left( {\dfrac{{\sin C}}{{\sin A\sin B}}} \right)\left( {\dfrac{{\sin A}}{{\sin B\sin C}}} \right)\left( {\dfrac{{\sin B}}{{\sin C\sin A}}} \right)$
Now on canceling the same terms as the numerators and denominators have the same term, so we get
$\Rightarrow \dfrac{1}{{\sin A\sin B\sin C}}$
And as we have already seen in the formula $\dfrac{1}{{\sin \theta }} = \cos ec \theta$ , so by using this we get
$\Rightarrow \cos ecA{\text{ cosecB cosecC}}$
Therefore, the value $\left( {\cot A + \cot B} \right)\left( {\cot B + \cot C} \right)\left( {\cot C + \cot A} \right)$ will be $\cos ecA{\text{ cosecB cosecC}}$

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

Note:
This type of question just needs one thing and it is the identities, as by using the identities and expanding the functions we can easily reduce the complexities, and also the chance of the error will be reduced. Also, we should note that the values given in the questions are very useful so we should always try to think of how we can use these values so that we can get to the result.