Question

Find the value of $x + y + z$ if ${\cot ^{ - 1}}x + {\cot ^{ - 1}}y + {\cot ^{ - 1}}z = \dfrac{\pi }{2}$
A) $xyz$
B) $xy + yz + zx$
C) $2xyz$
D) None of these

Answer 1

Hint- First, we will find the value of ${\cot ^{ - 1}}x$ in order to make our solution simple. Then we will replace that value in the given equation i.e. ${\cot ^{ - 1}}x + {\cot ^{ - 1}}y + {\cot ^{ - 1}}z = \dfrac{\pi }{2}$. We will also use the property of ($\tan( \pi - \theta) = -\tan(\theta) $).

Complete step-by-step answer:
The equation given to us by the question is: ${\cot ^{ - 1}}x + {\cot ^{ - 1}}y + {\cot ^{ - 1}}z = \dfrac{\pi }{2}$.
Now, as we all know that ${\cot ^{ - 1}}x + {\tan ^{ - 1}}x = \dfrac{\pi }{2}$, we will find out the value of ${\cot ^{ - 1}}x$ i.e. ${\cot ^{ - 1}}x = \dfrac{\pi }{2} - {\tan ^{ - 1}}x$. Now, we will replace this value of ${\cot ^{ - 1}}x$, in the equation given to us, we get-
$ \to \dfrac{\pi }{2} - {\tan ^{ - 1}}x + \dfrac{\pi }{2} - {\tan ^{ - 1}}y + \dfrac{\pi }{2} - {\tan ^{ - 1}}z = \dfrac{\pi }{2}$
Cancelling $\dfrac{\pi }{2}$ from the above equation we will have simplified equation as-
$ \to {\tan ^{ - 1}}x + {\tan ^{ - 1}}y + {\tan ^{ - 1}}z = \pi $
Taking ${\tan ^{ - 1}}z$ to the right side of the equation we get-
$ \to {\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi - {\tan ^{ - 1}}z$
Multiplying the above equation by $\tan $, we will get-
$ \to \tan \left( {{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y} \right) = \tan \left( {\pi - {{\tan }^{ - 1}}z} \right)$
By applying the property of $\tan \left( {\pi - \theta } \right) = - \tan \theta $ and the property of ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)$ in the above equation we will get-
$
   \to \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)} \right] = - \tan \left( {{{\tan }^{ - 1}}z} \right) \\
    \\
   \Rightarrow \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)} \right] = - z \\
    \\
   \Rightarrow \dfrac{{x + y}}{{1 - xy}} = - z \\
    \\
   \Rightarrow x + y = - z + xyz \\
    \\
   \Rightarrow x + y + z = xyz \\
$
Thus, we get the value of $x + y + z$ to be $xyz$
Hence, above option A is correct.

Note: Don’t forget to use the properties. Replacing some values by using properties will make the answer simple. The properties like ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)$ and $\tan \pi - \theta = - \tan \theta $ should always be in your mind.