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Question: If \( y = {\cot ^{ - 1}}\left[ {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right] \) find \( \df...

If y=cot⁡−1[cos⁡x−sin⁡xcos⁡x+sin⁡x]y = {\cot ^{ - 1}}\left[ {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right] find dydx\dfrac{{dy}}{{dx}} .

Explanation

Solution

Hint : Here, simplify the given trigonometric function, to make it in terms of tangent or cotangent. After that, simplify the inverse trigonometric function to obtain a simple value of function y in terms of x. Then differentiate with respect to x to get dydx\dfrac{{dy}}{{dx}} .

Complete step by step solution:
Given y=cot⁡−1[cos⁡x−sin⁡xcos⁡x+sin⁡x]y = {\cot ^{ - 1}}\left[ {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right]
Dividing numerator and denominator by cos x, we get
⇒y=cot⁡−1[cos⁡x−sin⁡xcos⁡xcos⁡x+sin⁡xcos⁡x]\Rightarrow y = {\cot ^{ - 1}}\left[ {\dfrac{{\dfrac{{\cos x - \sin x}}{{\cos x}}}}{{\dfrac{{\cos x + \sin x}}{{\cos x}}}}} \right]
On separating the trigonometric terms
⇒y=cot⁡−1[cos⁡xcos⁡x−sin⁡xcos⁡xcos⁡xcos⁡x+sin⁡xcos⁡x]\Rightarrow y = {\cot ^{ - 1}}\left[ {\dfrac{{\dfrac{{\cos x}}{{\cos x}} - \dfrac{{\sin x}}{{\cos x}}}}{{\dfrac{{\cos x}}{{\cos x}} + \dfrac{{\sin x}}{{\cos x}}}}} \right]
Representing terms in tangent form
y=cot⁡−1[1−tan⁡x1+tan⁡x]=cot⁡−1[tan⁡(π4−x)]y = {\cot ^{ - 1}}\left[ {\dfrac{{1 - \tan x}}{{1 + \tan x}}} \right] = {\cot ^{ - 1}}\left[ {\tan \left( {\dfrac{\pi }{4} - x} \right)} \right] [∵1−tan⁡x1+tan⁡x=tan⁡(π4−x)]\left[ {\because \dfrac{{1 - \tan x}}{{1 + \tan x}} = \tan \left( {\dfrac{\pi }{4} - x} \right)} \right]
Representing tangent and cotangent
\Rightarrow y = {\cot ^{ - 1}}\left[ {\cot \left\\{ {\dfrac{\pi }{2} - \left( {\dfrac{\pi }{4} - x} \right)} \right\\}} \right]
On simplifying
⇒y=cot⁡−1[cot⁡(π4+x)]=π4+x\Rightarrow y = {\cot ^{ - 1}}\left[ {\cot \left( {\dfrac{\pi }{4} + x} \right)} \right] = \dfrac{\pi }{4} + x
Now, we have y=π4+xy = \dfrac{\pi }{4} + x
Differentiating both sides with respect to x, we get
⇒dydx=1\Rightarrow \dfrac{{dy}}{{dx}} = 1

Note : In the above problem we can see that for differentiation first we have to apply the inverse formula of differentiation, then quotient rule then trigonometric differentiation and simplify the obtained result using many formulas. But we have first simplified the function and get linear relation between x and y and differentiation can be easily done and no simplification required. For all type of inverse trigonometric and trigonometric function first simplify using formulas and rules then differentiate to get the result with any complex calculations.