Question

Find the value of $\cot \left( {{{\tan }^{ - 1}}\alpha + {{\cot }^{ - 1}}\alpha } \right)$.

Answer 1

Hint: Here, we carry out simplification using an inverse trigonometric identity to find the value.

Complete step-by-step answer:
We have to find the value of $\cot \left( {{{\tan }^{ - 1}}\alpha + {{\cot }^{ - 1}}\alpha } \right)$
Now, we know that
$ta{n^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}$

Putting the above value in given question, we get
$ = \cot \left( {{{\tan }^{ - 1}}\alpha + {{\cot }^{ - 1}}\alpha } \right)$
$ = \cot \left( {\dfrac{\pi }{2}} \right)$
$ = \cot \left( {\dfrac{{{{180}^ \circ }}}{2}} \right) = \cot \left( {{{90}^ \circ }} \right) = 0$

Note: These types of questions can be solved if a student knows all the inverse trigonometric function identities. The value of trigonometric functions of standard angles must be remembered to arrive at the solution faster.