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Question: If\(y = {\tan ^{ - 1}}\sqrt {\dfrac{{a - x}}{{a + x}}} \), then\(\dfrac{{dy}}{{dx}} = \) A. \({\c...

Ify=tan⁡−1a−xa+xy = {\tan ^{ - 1}}\sqrt {\dfrac{{a - x}}{{a + x}}} , thendydx=\dfrac{{dy}}{{dx}} =
A. cos⁡−1(xa){\cos ^{ - 1}}\left( {\dfrac{x}{a}} \right)
B. −cos⁡−1(xa) - {\cos ^{ - 1}}\left( {\dfrac{x}{a}} \right)
C. 12cos⁡−1(xa)\dfrac{1}{2}{\cos ^{ - 1}}\left( {\dfrac{x}{a}} \right)
D. None of these

Explanation

Solution

First, we need to analyze the given information so that we can able to solve the problem. Generally, in Mathematics, the trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
Here we are asked to find the derivative ofy=tan⁡−1a−xa+xy = {\tan ^{ - 1}}\sqrt {\dfrac{{a - x}}{{a + x}}} .
We need to apply the appropriate trigonometric identities and derivative formula to obtain the required answer
Formula to be used:
a) The trigonometric identities that are used to solve the given problem are as follows.
1−cos⁡2θ=2sin⁡2θ1 - \cos 2\theta = 2{\sin ^2}\theta
1+cos⁡2θ=2cos⁡2θ1 + \cos 2\theta = 2{\cos ^2}\theta
sin⁡θcos⁡θ=tan⁡θ\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta
tan⁡−1tan⁡x=x{\tan ^{ - 1}}\tan x = x
b)ddx(cos⁡−1x)=−11−x2\dfrac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right) = \dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}

Complete step by step answer:
It is given thaty=tan⁡−1a−xa+xy = {\tan ^{ - 1}}\sqrt {\dfrac{{a - x}}{{a + x}}}
Let us putx=acos⁡2θx = a\cos 2\theta in the given equation.
x=acos⁡2θ⇒2θ=cos⁡−1(xa)x = a\cos 2\theta \Rightarrow 2\theta = {\cos ^{ - 1}}\left( {\dfrac{x}{a}} \right)
⇒θ=12cos⁡−1(xa)\Rightarrow \theta = \dfrac{1}{2}{\cos ^{ - 1}}\left( {\dfrac{x}{a}} \right) ……(1)\left( 1 \right)
Now, let us putx=acos⁡2θx = a\cos 2\theta in the given equation.
Hence, we gety=tan⁡−1a−acos⁡2θa+acos⁡2θy = {\tan ^{ - 1}}\sqrt {\dfrac{{a - a\cos 2\theta }}{{a + a\cos 2\theta }}}
=tan⁡−1a(1−cos⁡2θ)a(1+cos⁡2θ)= {\tan ^{ - 1}}\sqrt {\dfrac{{a\left( {1 - \cos 2\theta } \right)}}{{a\left( {1 + \cos 2\theta } \right)}}}
=tan⁡−11−cos⁡2θ1+cos⁡2θ= {\tan ^{ - 1}}\sqrt {\dfrac{{1 - \cos 2\theta }}{{1 + \cos 2\theta }}}
Now, we shall apply the formula1−cos⁡2θ=2sin⁡2θ1 - \cos 2\theta = 2{\sin ^2}\theta and1+cos⁡2θ=2cos⁡2θ1 + \cos 2\theta = 2{\cos ^2}\theta in the above equation.
Hence, we gety=tan⁡−12sin⁡2θ2cos⁡2θy = {\tan ^{ - 1}}\sqrt {\dfrac{{2{{\sin }^2}\theta }}{{2{{\cos }^2}\theta }}}
y=tan⁡−1sin⁡2θcos⁡2θy = {\tan ^{ - 1}}\sqrt {\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}
=tan⁡−1(sin⁡θcos⁡θ)2= {\tan ^{ - 1}}\sqrt {{{\left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right)}^2}}
=tan⁡−1sin⁡θcos⁡θ= {\tan ^{ - 1}}\dfrac{{\sin \theta }}{{\cos \theta }}
=tan⁡−1tan⁡θ= {\tan ^{ - 1}}\tan \theta (Here we applied the trigonometric identitysin⁡θcos⁡θ=tan⁡θ\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta )
=θ= \theta (Here we appliedtan⁡−1tan⁡x=x{\tan ^{ - 1}}\tan x = x )
Now, we need to substitute the equation(1)\left( 1 \right)in the above equation.
⇒y=12cos⁡−1(xa)\Rightarrow y = \dfrac{1}{2}{\cos ^{ - 1}}\left( {\dfrac{x}{a}} \right) ………(2)\left( 2 \right)
Here in this question, we are asked to calculate the derivative ofyy (i.e.dydx\dfrac{{dy}}{{dx}} )
Hence, we need to differentiate(2)\left( 2 \right)with respect toxx .
That isdydx=ddx(12cos⁡−1(xa))\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{x}{a}} \right)} \right)
=12ddx(cos⁡−1(xa))= \dfrac{1}{2}\dfrac{d}{{dx}}\left( {{{\cos }^{ - 1}}\left( {\dfrac{x}{a}} \right)} \right)
=12×−11−(xa)2×1a= \dfrac{1}{2} \times \dfrac{{ - 1}}{{\sqrt {1 - {{\left( {\dfrac{x}{a}} \right)}^2}} }} \times \dfrac{1}{a}
(Here we applied ddx(cos⁡−1x)=−11−x2\dfrac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right) = \dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }} )
=12×−1a2−x2a2×1a= \dfrac{1}{2} \times \dfrac{{ - 1}}{{\sqrt {\dfrac{{{a^2} - {x^2}}}{{{a^2}}}} }} \times \dfrac{1}{a}
=12×−a2a2−x2×1a= \dfrac{1}{2} \times \dfrac{{ - \sqrt {{a^2}} }}{{\sqrt {{a^2} - {x^2}} }} \times \dfrac{1}{a}
=12×−aa2−x2×1a= \dfrac{1}{2} \times \dfrac{{ - a}}{{\sqrt {{a^2} - {x^2}} }} \times \dfrac{1}{a}
=12×−1a2−x2= \dfrac{1}{2} \times \dfrac{{ - 1}}{{\sqrt {{a^2} - {x^2}} }}
=−12a2−x2= \dfrac{{ - 1}}{{2\sqrt {{a^2} - {x^2}} }}
Hence, dydx=−12a2−x2\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{2\sqrt {{a^2} - {x^2}} }}

So, the correct answer is “Option D”.

Note: If we are asked to calculate the answer that contains the value of a trigonometric expression, we need to first analyze the given problem where we are able to apply the trigonometric identities.
Here, we have applied some trigonometric identities/formulae and derivative formulae that are needed to know to obtain the desired answer. Hence, we gotdydx=−12a2−x2\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{2\sqrt {{a^2} - {x^2}} }}.