Question

If y = {\cot ^{ - 1}}\left\\{ {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right\\} find $\dfrac{{dy}}{{dx}}$ ?

Answer 1

Hint : Inverse trigonometric functions are simply defined as the inverse functions of the basic trigonometric functions which are sine, cosine, tangent, cotangent, secant, and cosecant functions. The given function is trigonometric with respect to inverse function, in which to find $\dfrac{{dy}}{{dx}}$ i.e., differentiating with respect to x, we must know all the basic inverse relations i.e., formulas of trigonometric identity functions and differentiate with respect to x.

Complete step by step solution:
Given,
y = {\cot ^{ - 1}}\left\\{ {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right\\}
Divide both numerator and denominator by cosx as:
\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\\}
We know that, $\dfrac{{\sin x}}{{\cos x}} = \tan x$ , on further simplification of terms, we get:
\Rightarrow y = {\cot ^{ - 1}}\left\\{ {\dfrac{{1 - \tan x}}{{1 + \tan x}}} \right\\}
We know that, $\tan \dfrac{\pi }{4} = 1$ , hence applying this we get:
\Rightarrow y = {\cot ^{ - 1}}\left\\{ {\dfrac{{\tan \dfrac{\pi }{4} - \tan x}}{{\tan \dfrac{\pi }{4} + \tan x}}} \right\\}
Now, combining the common terms we have:
\Rightarrow y = {\cot ^{ - 1}}\left\\{ {\tan \left( {\dfrac{\pi }{4} - x} \right)} \right\\}
Evaluating with respect to cot we get:
\Rightarrow y = {\cot ^{ - 1}}\left\\{ {\cot \left( {\dfrac{\pi }{2} - \left( {\dfrac{\pi }{4} - x} \right)} \right)} \right\\}
\Rightarrow y = {\cot ^{ - 1}}\left\\{ {\cot \left( {\dfrac{\pi }{4} + x} \right)} \right\\}
Simplifying the common functions, we get:
$\Rightarrow y = \dfrac{\pi }{4} + x$
Differentiating both sides with respect to x we have:
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{\pi }{4} + x} \right)$
Hence, differentiating the terms we get:
$\Rightarrow \dfrac{{dy}}{{dx}} = 1 + 0$
$\Rightarrow \dfrac{{dy}}{{dx}} = 1$
Therefore,
\dfrac{d}{{dx}}\left( {{{\cot }^{ - 1}}\left\\{ {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right\\}} \right) = 1
So, the correct answer is “1”.

Note : The key point to solve any trigonometric function is that we must know all the formulas with respect to the related question asked as it seems easy to solve the question, we must note the chart of all related functions with respect to the given equation, and here are some of the formulas and relations to be noted while solving: $\dfrac{{\sin x}}{{\cos x}} = \tan x$ , $\tan \dfrac{\pi }{4} = 1$ .