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

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

Explanation

Solution

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⁡(tan⁡−1α+cot⁡−1α)\cot \left( {{{\tan }^{ - 1}}\alpha + {{\cot }^{ - 1}}\alpha } \right)
Now, we know that
tan−1x+cot⁡−1x=π2ta{n^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}

Putting the above value in given question, we get
=cot⁡(tan⁡−1α+cot⁡−1α)= \cot \left( {{{\tan }^{ - 1}}\alpha + {{\cot }^{ - 1}}\alpha } \right)
=cot⁡(π2)= \cot \left( {\dfrac{\pi }{2}} \right)
=cot⁡(180∘2)=cot⁡(90∘)=0= \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.