Question

Using derivative, prove that: \[{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}\].

Answer 1

Hint: Here we have to prove the given trigonometric identity \[{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}\] using derivative. We can first differentiate the given function with respect to x and simplify it. We will get \[f'\left( x \right)=0\], we should know when \[f'\left( x \right)=0\], \[f\left( x \right)\] is a constant function. We can then take x = 0 and substitute it in the given expression to get the value of the given function. We can then check for the left and the right-hand side to prove the problem.

Complete step by step solution:
Here we have to prove \[{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}\] using derivative.
We can now write the left-hand side of the given expression as,
\[\Rightarrow f\left( x \right)={{\tan }^{-1}}x+{{\cot }^{-1}}x\]…… (1)
We can now differentiate the above function with respect to x, we get
\[\Rightarrow f'\left( x \right)=\dfrac{d}{dx}\left( {{\tan }^{-1}}x+{{\cot }^{-1}}x \right)\]
We can now find the derivative and simplify it, we get
\[\begin{align}
  & \Rightarrow f'\left( x \right)=\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)+\dfrac{d}{dx}\left( {{\cot }^{-1}}x \right) \\
 & \Rightarrow f'\left( x \right)=\dfrac{1}{1+{{x}^{2}}}-\dfrac{1}{1+{{x}^{2}}}=0 \\
\end{align}\]
We can see that \[f'\left( x \right)=0\].
Since \[f'\left( x \right)=0\], \[f\left( x \right)\] is a constant function.
Let \[f\left( x \right)=k\]
We can now take that, for any value of x, \[f\left( x \right)=k\]
Let x = 0, then \[f\left( 0 \right)=k\]
We can now substitute the value of x in (1), we get
\[\Rightarrow f\left( 0 \right)={{\tan }^{-1}}0+{{\cot }^{-1}}0=0+\dfrac{\pi }{2}\]
\[\because {{\tan }^{-1}}0=0,{{\cot }^{-1}}0=\dfrac{\pi }{2}\]
Therefore, the value of \[f\left( x \right)=k=\dfrac{\pi }{2}\] ….. (2)
We can see that from (1) and (2)
\[{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}\].
Hence proved.

Note: We should always remember the derivative formulas to solve these types of problems, we should remember that the derivative of \[{{\tan }^{-1}}x=\dfrac{1}{1+{{x}^{2}}},{{\cot }^{-1}}x=\dfrac{-1}{1+{{x}^{2}}}\]. We should also remember that differentiating the given function if \[f'\left( x \right)=0\], \[f\left( x \right)\] is a constant function.
We should also know some of the trigonometric degree values to be substituted such as
\[{{\tan }^{-1}}0=0,{{\cot }^{-1}}0=\dfrac{\pi }{2}\].