Question

If $A + B + C = \pi$ then prove that $\cot A + \cot B + \cot C = \cot A\cot B\cot C + \csc A\csc B\csc C$ .

Answer 1

We will use the ratios of trigonometry and some additional trigonometric formulas.in this problem we will take $\csc$ function from right side to left side since $\csc$ and $\cot$ have $\sin$ function in denominator. $\csc A\csc B\csc C$ will help in having $\sin$ function in product form.
$\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
$\csc \theta = \dfrac{1}{{\sin \theta }}$
$\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta$

Complete step-by-step answer:
We are given that
$\cot A + \cot B + \cot C = \cot A\cot B\cot C + \csc A\csc B\csc C$
We will modify tis statement for our convenience as
$\cot A + \cot B + \cot C - \csc A\csc B\csc C = \cot A\cot B\cot C$
Now we will start with the LHS.
$\Rightarrow \cot A + \cot B + \cot C - \csc A\csc B\csc C$
Using the ratios mentioned in hint as $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\& \csc \theta = \dfrac{1}{{\sin \theta }}$ we get
$\Rightarrow \dfrac{{\cos A}}{{\sin A}} + \dfrac{{\cos B}}{{\sin B}} + \dfrac{{\cos C}}{{\sin C}} - \dfrac{1}{{\sin A\sin B\sin C}}$
Now we will find the LCM of the first two terms. And the LCM is $\sin A\sin B$
$\Rightarrow \dfrac{{\cos A\sin B}}{{\sin A\sin B}} + \dfrac{{\cos B\sin A}}{{\sin A\sin B}} + \dfrac{{\cos C}}{{\sin C}} - \dfrac{1}{{\sin A\sin B\sin C}}$
$\Rightarrow \dfrac{{\cos A\sin B + \cos B\sin A}}{{\sin A\sin B}} + \dfrac{{\cos C}}{{\sin C}} - \dfrac{1}{{\sin A\sin B\sin C}}$

Now the first two terms give the formula of $\sin (A + B) = \sin A\cos B + \cos A\sin B$
$\Rightarrow \dfrac{{\sin \left( {A + B} \right)}}{{\sin A\sin B}} + \dfrac{{\cos C}}{{\sin C}} - \dfrac{1}{{\sin A\sin B\sin C}}$
Now we are given that

A + B + C = \pi \\\ \Rightarrow A + B = \pi - C \\\ \end{gathered} $$ So $$\sin \left( {A + B} \right) = \sin \left( {\pi - C} \right) = \sin C$$ Then replacing the numerator of the first term with $$\sin C$$ we get $$ \Rightarrow \dfrac{{\sin C}}{{\sin A\sin B}} + \dfrac{{\cos C}}{{\sin C}} - \dfrac{1}{{\sin A\sin B\sin C}}$$ Now taking LCM of all the three terms we get $$ \Rightarrow \dfrac{{\sin C\sin C}}{{\sin A\sin B\sin C}} + \dfrac{{\cos C\sin A\sin B}}{{\sin C\sin A\sin B}} - \dfrac{1}{{\sin A\sin B\sin C}}$$ Now since denominators are same we can write the above step as, $$ \Rightarrow \left[ {{{\sin }^2}C + \cos C\sin A\sin B - 1} \right]\dfrac{1}{{\sin A\sin B\sin C}}$$ Now from above step we recall a famous trigonometric identity $${\cos ^2}\theta + {\sin ^2}\theta = 1$$ But we will modify it as per our problem demands, $$ \Rightarrow \left[ {\cos C\sin A\sin B + {{\sin }^2}C - 1} \right]\dfrac{1}{{\sin A\sin B\sin C}}$$ $$ \Rightarrow \left[ {\cos C\sin A\sin B - {{\cos }^2}C} \right]\dfrac{1}{{\sin A\sin B\sin C}}$$ Hope upto this we are very clear. Let’s proceed further! Now taking $$\cos C$$ common from both the terms we get, $$ \Rightarrow \left[ {\sin A\sin B - \cos C} \right]\dfrac{{\cos C}}{{\sin A\sin B\sin C}}$$ Now this is the important step of the problem. If you observe we have $$\cos C$$ in the numerator and $$\sin C$$ in the denominator of the last term, $$ \Rightarrow \left[ {\sin A\sin B - \cos C} \right]\dfrac{{\cos C}}{{\sin A\sin B\sin C}}$$ This will come out as $$\cot C$$ . But we have two more ratios to find. So we need $$\cos A\cos B$$ in the numerator .So let’s get them! Now $$\cos C$$ can be written as, $$\cos \left( {\pi - \left( {A + B} \right)} \right)$$ Replace this value in the above step we get, $$ \Rightarrow \left[ {\sin A\sin B - \cos \left( {\pi - \left( {A + B} \right)} \right)} \right]\dfrac{{\cos C}}{{\sin A\sin B\sin C}}$$ And $$\cos \left( {\pi - \left( {A + B} \right)} \right) = - \cos \left( {A + B} \right)$$ $$ \Rightarrow \left[ {\sin A\sin B - \left( { - \cos \left( {A + B} \right)} \right)} \right]\dfrac{{\cos C}}{{\sin A\sin B\sin C}}$$ $$ \Rightarrow \left[ {\sin A\sin B + \cos \left( {A + B} \right)} \right]\dfrac{{\cos C}}{{\sin A\sin B\sin C}}$$ Now we know that $$\cos \left( {A + B} \right) = \cos A\cos B + \sin A\sin B$$ $$ \Rightarrow \left[ {\sin A\sin B + \cos A\cos B - \sin A\sin B} \right]\dfrac{{\cos C}}{{\sin A\sin B\sin C}}$$ The terms with sin function will be cancelled, $$ \Rightarrow \dfrac{{\cos A\cos B\cos C}}{{\sin A\sin B\sin C}}$$ Here we come to end of our proof that, $$ \Rightarrow \cot A\cot B\cot C$$ → Since $$\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta $$ $$ = RHS$$. Hence proved. **Note:** These types of problems need practice because the more we practice the more we come to know how to modify the trigonometric identities. Generally students directly start with the LHS or RHS of the statement but some problems like this need some modifications then only we can solve further.