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Question: If \[y = {\tan ^{ - 1}}\left( {\dfrac{{a\cos x - b\sin x}}{{b\cos x + a\sin x}}} \right)\] then \[\d...

If y=tan⁡−1(acos⁡x−bsin⁡xbcos⁡x+asin⁡x)y = {\tan ^{ - 1}}\left( {\dfrac{{a\cos x - b\sin x}}{{b\cos x + a\sin x}}} \right) then dydx=\dfrac{{dy}}{{dx}} = ?
A) ab\dfrac{a}{b}
B) −ba\dfrac{{ - b}}{a}
C) 11
D) −1 - 1

Explanation

Solution

We will solve this question by using the inverse trigonometric formula tan⁡−1x−tan⁡−1y{\tan ^{ - 1}}x - {\tan ^{ - 1}}y as shown below:
tan⁡−1x−tan⁡−1y=tan⁡−1[(x−y)(1+xy)]{\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {x - y} \right)}}{{\left( {1 + xy} \right)}}} \right] , where xx and yy are two variables.

Complete step-by-step solution:
Step 1: It is given that y=tan⁡−1(acos⁡x−bsin⁡xbcos⁡x+asin⁡x)y = {\tan ^{ - 1}}\left( {\dfrac{{a\cos x - b\sin x}}{{b\cos x + a\sin x}}} \right). By dividing the numerator and denominator of the RHS side of the expression by bcos⁡xb\cos x, we get:
y=tan⁡−1(acos⁡x−bsin⁡xbcos⁡xbcos⁡x+asin⁡xbcos⁡x)y = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{a\cos x - b\sin x}}{{b\cos x}}}}{{\dfrac{{b\cos x + a\sin x}}{{b\cos x}}}}} \right)
Now by dividing and numerator and denominator part of the RHS side of the expression, we get:
⇒y=tan⁡−1(ab−sin⁡xcos⁡x1+absin⁡xcos⁡x)\Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{a}{b} - \dfrac{{\sin x}}{{\cos x}}}}{{1 + \dfrac{a}{b}\dfrac{{\sin x}}{{\cos x}}}}} \right)
By putting sin⁡xcos⁡x=tan⁡x\dfrac{{\sin x}}{{\cos x}} = \tan x in the above expression y=tan⁡−1(ab−sin⁡xcos⁡x1+absin⁡xcos⁡x)y = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{a}{b} - \dfrac{{\sin x}}{{\cos x}}}}{{1 + \dfrac{a}{b}\dfrac{{\sin x}}{{\cos x}}}}} \right) , we get:
⇒y=tan⁡−1(ab−tan⁡x1+abtan⁡x)\Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{a}{b} - \tan x}}{{1 + \dfrac{a}{b}\tan x}}} \right) ……………. (1)
Step 2: Now, by applying the formula of tan⁡−1x−tan⁡−1y=tan⁡−1[(x−y)(1+xy)]{\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {x - y} \right)}}{{\left( {1 + xy} \right)}}} \right] in the expression (1), we get:
⇒y=tan⁡−1(ab)−tan⁡−1(tan⁡x)\Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{a}{b}} \right) - {\tan ^{ - 1}}\left( {\tan x} \right) , where x=abx = \dfrac{a}{b} and y=tan⁡xy = \tan x.
We know that tan⁡−1(tan⁡x)=x{\tan ^{ - 1}}\left( {\tan x} \right) = x , (∵tan⁡−1x=1tan⁡x)\left( {\because {{\tan }^{ - 1}}x = \dfrac{1}{{\tan x}}} \right) , substituting this value in the above expression y=tan⁡−1(ab)−tan⁡−1(tan⁡x)y = {\tan ^{ - 1}}\left( {\dfrac{a}{b}} \right) - {\tan ^{ - 1}}\left( {\tan x} \right), we get:
⇒y=tan⁡−1(ab)−x\Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{a}{b}} \right) - x ………… (2)
Step 3: Differentiating the expression (2), for xx, we get:
⇒dydx=0−(1)\Rightarrow \dfrac{{dy}}{{dx}} = 0 - \left( 1 \right) , (∵tan⁡−1ab=constant)\left( {\because {{\tan }^{ - 1}}\dfrac{a}{b} = {\text{constant}}} \right)
By simplifying into the RHS side of the above expression, we get:
⇒dydx=−1\Rightarrow \dfrac{{dy}}{{dx}} = - 1
Therefore, if y=tan⁡−1(acos⁡x−bsin⁡xbcos⁡x+asin⁡x)y = {\tan ^{ - 1}}\left( {\dfrac{{a\cos x - b\sin x}}{{b\cos x + a\sin x}}} \right), then dydx=−1\dfrac{{dy}}{{dx}} = - 1.

Option D is the correct answer.

Note: While solving these types of questions students should remember some formulas to make the calculation easy. For example, we have used the formula of tan⁡−1x−tan⁡−1y=tan⁡−1[(x−y)(1+xy)]{\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {x - y} \right)}}{{\left( {1 + xy} \right)}}} \right] , explanation of which is as given below for your better understanding:
To prove tan⁡−1x−tan⁡−1y=tan⁡−1[(x−y)(1+xy)]{\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {x - y} \right)}}{{\left( {1 + xy} \right)}}} \right] , let tan⁡−1x=A{\tan ^{ - 1}}x = {\text{A}},tan⁡−1y=B{\tan ^{ - 1}}y = {\text{B}}.
As we know that tan⁡(A−B)=tan⁡A−tan⁡B1−tan⁡Atan⁡B\tan \left( {{\text{A}} - {\text{B}}} \right) = \dfrac{{\tan {\text{A}} - \tan {\text{B}}}}{{1 - \tan {\text{A}}\tan {\text{B}}}} , so by substituting the values of tan⁡−1x=A{\tan ^{ - 1}}x = {\text{A}},tan⁡−1y=B{\tan ^{ - 1}}y = {\text{B}} in this formula, we get:
⇒tan⁡(tan⁡−1x−tan⁡−1y)=tan⁡(tan⁡−1x)−tan⁡(tan⁡−1y)1+tan⁡(tan⁡−1x)tan⁡(tan⁡−1y)\Rightarrow \tan \left( {{{\tan }^{ - 1}}x - {{\tan }^{ - 1}}y} \right) = \dfrac{{\tan \left( {{{\tan }^{ - 1}}x} \right) - \tan \left( {{{\tan }^{ - 1}}y} \right)}}{{1 + \tan \left( {{{\tan }^{ - 1}}x} \right)\tan \left( {{{\tan }^{ - 1}}y} \right)}}
From here, we can write tan⁡(tan⁡−1x)=x\tan \left( {{{\tan }^{ - 1}}x} \right) = x and tan⁡(tan⁡−1y)=y\tan \left( {{{\tan }^{ - 1}}y} \right) = y because f(f−1a)=af\left( {{f^{ - 1}}a} \right) = a, where ff is known as any function. By substituting these values in the above expression, we get:
⇒tan⁡(tan⁡−1x−tan⁡−1y)=x−y1+xy\Rightarrow \tan \left( {{{\tan }^{ - 1}}x - {{\tan }^{ - 1}}y} \right) = \dfrac{{x - y}}{{1 + xy}}
Now by bringing tan⁡\tan from the LHS side to the RHS side of the expression we get:
⇒(tan⁡−1x−tan⁡−1y)=tan⁡−1x−y1+xy\Rightarrow \left( {{{\tan }^{ - 1}}x - {{\tan }^{ - 1}}y} \right) = {\tan ^{ - 1}}\dfrac{{x - y}}{{1 + xy}}
It is proved that tan⁡−1x−tan⁡−1y=tan⁡−1x−y1+xy{\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\dfrac{{x - y}}{{1 + xy}}