Solveeit Logo
Login

Question

Question: State whether the given statement is true or false \[{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\p...

State whether the given statement is true or false
tan⁡−1x+cot⁡−1x=π2,x∈R{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2},x \in R
A. TRUE
B. FALSE

Explanation

Solution

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{f^{ - 1}}(f(x)) = x.
  • Complementary angles means that the angles have sum of 90∘{90^ \circ }i.e. for the functions ‘tan’ and ‘cot’ tan⁡(π2−θ)=cot⁡θ\tan (\dfrac{\pi }{2} - \theta ) = \cot \theta andcot⁡(π2−θ)=tan⁡θ\cot (\dfrac{\pi }{2} - \theta ) = \tan \theta

Complete step by step solution:
We are given the equation tan⁡−1x+cot⁡−1x=π2,x∈R{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2},x \in R
We consider one part of the LHS as a variable.
Let tan⁡−1x=y{\tan ^{ - 1}}x = y.........… (1)
Now take same function on both sides of the equation
⇒tan⁡(tan⁡−1x)=tan⁡(y)\Rightarrow \tan \left( {{{\tan }^{ - 1}}x} \right) = \tan \left( y \right)
Since we know f−1(f(x))=x{f^{ - 1}}(f(x)) = x
⇒x=tan⁡y\Rightarrow x = \tan y............… (2)
Since tan⁡\tan and cot⁡\cot are complementary angles
We can write tan⁡y=cot⁡(π2−y)\tan y = \cot \left( {\dfrac{\pi }{2} - y} \right)
Substitute the value of tan⁡y\tan yin equation (2)
⇒x=cot⁡(π2−y)\Rightarrow x = \cot \left( {\dfrac{\pi }{2} - y} \right)
Now take inverse function of cot⁡\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{f^{ - 1}}(f(x)) = x
⇒cot⁡−1(x)=(π2−y)\Rightarrow {\cot ^{ - 1}}\left( x \right) = \left( {\dfrac{\pi }{2} - y} \right)
Shift ‘y’ to LHS of the equation
⇒cot⁡−1x+y=π2\Rightarrow {\cot ^{ - 1}}x + y = \dfrac{\pi }{2}
Substitute the value of ‘y’ back from equation (1)
⇒cot⁡−1x+tan⁡−1x=π2\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⁡θ+cot⁡θ=90∘\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.