Question

State whether the given statement is true or false
\[{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2},x \in R\]
A. TRUE
B. FALSE

Answer 1

Hint: Here we check the equation is true or false by considering one term of the LHS as a variable and then applying the same function to that equation whichever inverse we have assumed as a variable. Use the concept of complementary angles to convert the angle into its complement. Again take the required inverse that cancels out the existing function in the equation.
* Inverse of any function when applied on the same function cancels out the function and the inverse, i.e. \[{f^{ - 1}}(f(x)) = x\].
* Complementary angles means that the angles have sum of \[{90^ \circ }\]i.e. for the functions ‘tan’ and ‘cot’ \[\tan (\dfrac{\pi }{2} - \theta ) = \cot \theta \]and\[\cot (\dfrac{\pi }{2} - \theta ) = \tan \theta \]

Complete step by step solution:
We are given the equation \[{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2},x \in R\]
We consider one part of the LHS as a variable.
Let \[{\tan ^{ - 1}}x = y\].........… (1)
Now take same function on both sides of the equation
\[ \Rightarrow \tan \left( {{{\tan }^{ - 1}}x} \right) = \tan \left( y \right)\]
Since we know \[{f^{ - 1}}(f(x)) = x\]
\[ \Rightarrow x = \tan y\]............… (2)
Since \[\tan \] and \[\cot \] are complementary angles
We can write \[\tan y = \cot \left( {\dfrac{\pi }{2} - y} \right)\]
Substitute the value of \[\tan y\]in equation (2)
\[ \Rightarrow x = \cot \left( {\dfrac{\pi }{2} - y} \right)\]
Now take inverse function of \[\cot \]on both sides of the equation
\[ \Rightarrow {\cot ^{ - 1}}\left( x \right) = {\cot ^{ - 1}}\left\{ {\cot \left( {\dfrac{\pi }{2} - y} \right)} \right\}\]
Since we know \[{f^{ - 1}}(f(x)) = x\]
\[ \Rightarrow {\cot ^{ - 1}}\left( x \right) = \left( {\dfrac{\pi }{2} - y} \right)\]
Shift ‘y’ to LHS of the equation
\[ \Rightarrow {\cot ^{ - 1}}x + y = \dfrac{\pi }{2}\]
Substitute the value of ‘y’ back from equation (1)
\[ \Rightarrow {\cot ^{ - 1}}x + {\tan ^{ - 1}}x = \dfrac{\pi }{2}\]
\[\therefore \]LHS\[ = \]RHS
Hence Proved
\[\therefore \]Equation given in the question is TRUE

\[\therefore \]Option A is correct.

Note: Students many times make mistake of taking complementary functions instead of complementary angles and they write \[\tan \theta + \cot \theta = {90^ \circ }\] which is wrong because by complementary functions we mean the functions have sum 90 and by complementary angles we mean sum of angles is 90. Keep in mind to cancel the inverse and the function both the inverse and the function should be the same, many students try to cancel the inverse of tan with the ‘cot’ function which is wrong.