Question

If \[y = {\tan ^{ - 1}}\left( {\dfrac{{a\cos x - b\sin x}}{{b\cos x + a\sin x}}} \right)\] then \[\dfrac{{dy}}{{dx}} = \]?
A) \[\dfrac{a}{b}\]
B) \[\dfrac{{ - b}}{a}\]
C) \[1\]
D) \[ - 1\]

Answer 1

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

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

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 \[{\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {x - y} \right)}}{{\left( {1 + xy} \right)}}} \right]\] , explanation of which is as given below for your better understanding:
To prove \[{\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {x - y} \right)}}{{\left( {1 + xy} \right)}}} \right]\] , let \[{\tan ^{ - 1}}x = {\text{A}}\],\[{\tan ^{ - 1}}y = {\text{B}}\].
As we know that \[\tan \left( {{\text{A}} - {\text{B}}} \right) = \dfrac{{\tan {\text{A}} - \tan {\text{B}}}}{{1 - \tan {\text{A}}\tan {\text{B}}}}\] , so by substituting the values of \[{\tan ^{ - 1}}x = {\text{A}}\],\[{\tan ^{ - 1}}y = {\text{B}}\] in this formula, we get:
\[ \Rightarrow \tan \left( {{{\tan }^{ - 1}}x - {{\tan }^{ - 1}}y} \right) = \dfrac{{\tan \left( {{{\tan }^{ - 1}}x} \right) - \tan \left( {{{\tan }^{ - 1}}y} \right)}}{{1 + \tan \left( {{{\tan }^{ - 1}}x} \right)\tan \left( {{{\tan }^{ - 1}}y} \right)}}\]
From here, we can write \[\tan \left( {{{\tan }^{ - 1}}x} \right) = x\] and \[\tan \left( {{{\tan }^{ - 1}}y} \right) = y\] because \[f\left( {{f^{ - 1}}a} \right) = a\], where \[f\] is known as any function. By substituting these values in the above expression, we get:
\[ \Rightarrow \tan \left( {{{\tan }^{ - 1}}x - {{\tan }^{ - 1}}y} \right) = \dfrac{{x - y}}{{1 + xy}}\]
Now by bringing \[\tan \]from the LHS side to the RHS side of the expression we get:
\[ \Rightarrow \left( {{{\tan }^{ - 1}}x - {{\tan }^{ - 1}}y} \right) = {\tan ^{ - 1}}\dfrac{{x - y}}{{1 + xy}}\]
It is proved that \[{\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\dfrac{{x - y}}{{1 + xy}}\]