Question

If ${{\cot }^{-1}}x+{{\cot }^{-1}}y + {{\cot }^{-1}}z =\dfrac{\pi }{2}$ then x + y + z equals ?

Answer 1

To do this question, first you need to take the cot inverse z to the right side. Then apply tan on both sides. So, now we can use the tan( A + B ) formula. $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$. Also you need to apply the formula $\tan \left( {{\cot }^{-1}}x \right)=\dfrac{1}{x}$. Then after simplifying all the equations, you get the final equation in terms of x + y + z. The value on the right side is the answer.

Complete step by step solution:
Here is the step wise solution.
The first step to do is to ${{\cot }^{-1}}z$ to the right side. Therefore, we get the equation as
${{\cot }^{-1}}x+{{\cot }^{-1}}y=\dfrac{\pi }{2}-{{\cot }^{-1}}z$.
The next step is to apply tan on both sides of the equation, so that we can apply the tan(A+B) formula. Then we get:
$\tan \left( {{\cot }^{-1}}x+{{\cot }^{-1}}y \right)=\tan \left( \dfrac{\pi }{2}-{{\cot }^{-1}}z \right)$.
Now we use the formula of tan(A + B). The formula is $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$. Also we need to apply the formula $\tan \left( {{\cot }^{-1}}x \right)=\dfrac{1}{x}$. So, now we get
$\Rightarrow \dfrac{\tan \left( {{\cot }^{-1}}x \right)+\tan \left( {{\cot }^{-1}}y \right)}{1-\tan \left( {{\cot }^{-1}}x \right)\tan \left( {{\cot }^{-1}}x \right)}=\cot \left( {{\cot }^{-1}}z \right)$
Simplifying this, we get
$\Rightarrow \dfrac{\dfrac{1}{x}+\dfrac{1}{y}}{1-\dfrac{1}{x}\dfrac{1}{y}}=z$
$\Rightarrow \dfrac{x+y}{xy-1}=z$
$\Rightarrow x+y=xyz-z$
$\Rightarrow x+y+z=xyz$
Therefore, as we can see , we get the final answer for the question as xyz.

Note: You need to remember the trigonometric identities. It will be easy to use all the formulas and solve this question. You could also apply cot instead of tan on both sides of the equation if you know the cot (A + B ) formula. Be careful while using the formulas.