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Question: If \(y = {\tan ^{ - 1}}\left( {\cot \left( {\dfrac{\pi }{2} - x} \right)} \right)\), then \(\dfrac{{...

If y=tan⁡−1(cot⁡(π2−x))y = {\tan ^{ - 1}}\left( {\cot \left( {\dfrac{\pi }{2} - x} \right)} \right), then dydx=\dfrac{{dy}}{{dx}} =

Explanation

Solution

Hint : Here, we cannot directly differentiate y=tan⁡−1(cot⁡(π2−x))y = {\tan ^{ - 1}}\left( {\cot \left( {\dfrac{\pi }{2} - x} \right)} \right). First of all, we will be using the relation
⇒cot⁡(π2−θ)=tan⁡θ ⇒tan⁡(π2−θ)=cot⁡θ   \Rightarrow \cot \left( {\dfrac{\pi }{2} - \theta } \right) = \tan \theta \\\ \Rightarrow \tan \left( {\dfrac{\pi }{2} - \theta } \right) = \cot \theta \;
And then tan⁡−1(tan⁡x)=x{\tan ^{ - 1}}\left( {\tan x} \right) = x. After using these relations, we can easily differentiate the given equation.

Complete step-by-step answer :
In this question, we are given an equation and we need to find its derivative, that is dydx\dfrac{{dy}}{{dx}}.
Given equation is: y=tan⁡−1(cot⁡(π2−x))y = {\tan ^{ - 1}}\left( {\cot \left( {\dfrac{\pi }{2} - x} \right)} \right) - - - - - - - - - - - - - - - - - - (1)
Now, we cannot differentiate the above equation directly. We need to use some mathematical formulas and relations in order to find the derivative of the given equation.
Now, first of all, we know that cot and tan are complementary to each other. So, we have the relation between them that
⇒cot⁡(π2−θ)=tan⁡θ ⇒tan⁡(π2−θ)=cot⁡θ   \Rightarrow \cot \left( {\dfrac{\pi }{2} - \theta } \right) = \tan \theta \\\ \Rightarrow \tan \left( {\dfrac{\pi }{2} - \theta } \right) = \cot \theta \;
So, therefore equation (1) becomes
⇒y=tan⁡−1(tan⁡x)\Rightarrow y = {\tan ^{ - 1}}\left( {\tan x} \right) - - - - - - - - - - (2)
Now, we know another relation that when any trigonometric ratio is multiplied by its inverse, then it will always be equal to its angle. Therefore, we have
⇒tan⁡−1(tan⁡x)=x\Rightarrow {\tan ^{ - 1}}\left( {\tan x} \right) = x
Therefore, putting this value in equation (2), we get
⇒y=tan⁡−1(tan⁡x)\Rightarrow y = {\tan ^{ - 1}}\left( {\tan x} \right)
⇒y=x\Rightarrow y = x - - - - - - - - - - (3)
Now, we can differentiate the equation directly.
Therefore, on differentiating equation (3), we get
⇒y=x ⇒dydx=ddx(x) ⇒dydx=1   \Rightarrow y = x \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( x \right) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = 1 \;
Hence, the derivative of y=tan⁡−1(cot⁡(π2−x))y = {\tan ^{ - 1}}\left( {\cot \left( {\dfrac{\pi }{2} - x} \right)} \right) is equal to 1.
So, the correct answer is “1”.

Note : Here, we can also find the derivative of y=tan⁡−1(cot⁡(π2−x))y = {\tan ^{ - 1}}\left( {\cot \left( {\dfrac{\pi }{2} - x} \right)} \right) using the above method.
⇒y=tan⁡−1(cot⁡(π2−x))\Rightarrow y = {\tan ^{ - 1}}\left( {\cot \left( {\dfrac{\pi }{2} - x} \right)} \right)
We know that, ddxtan⁡−1x=11+x2\dfrac{d}{{dx}}{\tan ^{ - 1}}x = \dfrac{1}{{1 + {x^2}}}
Therefore,
⇒dydx=ddxtan⁡−1(cot⁡(π2−x)) ⇒dydx=11+cot⁡2(π2−x)×ddxcot⁡(π2−x)×ddx(−x) ⇒dydx=1cos⁡ec2(π2−x)×(−cos⁡ec2(π2−x))(−1) ⇒dydx=(−1)(−1) ⇒dydx=1   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{\tan ^{ - 1}}\left( {\cot \left( {\dfrac{\pi }{2} - x} \right)} \right) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{1 + {{\cot }^2}\left( {\dfrac{\pi }{2} - x} \right)}} \times \dfrac{d}{{dx}}\cot \left( {\dfrac{\pi }{2} - x} \right) \times \dfrac{d}{{dx}}\left( { - x} \right) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\cos e{c^2}\left( {\dfrac{\pi }{2} - x} \right)}} \times \left( { - \cos e{c^2}\left( {\dfrac{\pi }{2} - x} \right)} \right)\left( { - 1} \right) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \left( { - 1} \right)\left( { - 1} \right) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = 1 \;