Question

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

Answer 1

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

& \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{aligned}$$ We can see that $$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}$$.