Question

If $\sin y = x\sin\left( {a + y} \right)$, and $\dfrac{{dy}}{{dx}} = \dfrac{A}{{\left[ {1 + {x^2} - 2x\cos a} \right]}}$. Then what is the value of $A$?
A. 2
B. $\cos a$
C. $\sin a$
D. None of these

Answer 1

Hint: Simplify the given trigonometric equation by using the trigonometric identity $\sin\left( {A + B} \right)$. Then divide both sides of the equation by $\cos y$ and get the equation of $\tan y$. In the end, differentiate the equation with respect to $x$ and solve it using the quotient rule to get the required answer.

Formula Used:
$\tan x = \dfrac{{\sin x}}{{\cos x}}$
$\sin\left( {A + B} \right) = \sin A\cos B + \cos A\sin B$
$\sin^{2}x + \cos^{2}x = 1$
$\dfrac{d}{{dx}}\tan^{ - 1}x = \dfrac{1}{{1 + {x^2}}} \cdot \dfrac{d}{{dx}}\left( x \right)$
Quotient rule of differentiation: $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{vu' - uv'}}{{{v^2}}}$

Complete step by step solution:
The given trigonometric equation is $\sin y = x\sin\left( {a + y} \right)$ and the differential equation is $\dfrac{{dy}}{{dx}} = \dfrac{A}{{\left[ {1 + {x^2} - 2x\cos a} \right]}}$.
Let’s simplify the trigonometric equation.
$\sin y = x\sin\left( {a + y} \right)$
Apply the identity $\sin\left( {A + B} \right) = \sin A\cos B + \cos A\sin B$.
$\sin y = x\left[ {\sin a \cos y + \cos a \sin y} \right]$
Divide both sides by $\cos y$.
$\dfrac{{\sin y}}{{\cos y}} = \dfrac{x}{{\cos y}}\left[ {\sin a \cos y + \cos a \sin y} \right]$
Simplify the above equation.
$\tan y = x\left[ {\dfrac{{\sin a \cos y}}{{\cos y}} + \dfrac{{\cos a \sin y}}{{\cos y}}} \right]$ [Since $\tan x = \dfrac{{\sin x}}{{\cos x}}$]
$ \Rightarrow $$\tan y = x\left[ {\sin a + \cos a \tan y} \right]$
$ \Rightarrow $$\tan y – x\cos a \tan y = x\sin a$
$ \Rightarrow $$\tan y\left( {1 – x\cos a} \right) = x\sin a$
Divide both sides by $\left( {1 – x\cos a} \right) $.
$\tan y = \dfrac{{x\sin a}}{{\left( {1 – x\cos a} \right)}}$
$ \Rightarrow $$y = \tan^{ - 1}\left( {\dfrac{{x\sin a}}{{1 – x\cos a}}} \right)$

Now differentiate the above equation with respect to $x$.
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\tan^{ - 1}\left( {\dfrac{{x\sin a}}{{1 – x\cos a}}} \right)$
Apply the derivative formula $\dfrac{d}{{dx}}\tan^{ - 1}x = \dfrac{1}{{1 + {x^2}}} \cdot \dfrac{d}{{dx}}\left( x \right)$.
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{1 + {{\left( {\dfrac{{x\sin a}}{{1 - x\cos a}}} \right)}^2}}} \cdot \dfrac{d}{{dx}} \left( {\dfrac{{x\sin a}}{{1 - x\cos a}}} \right)$
Simplify the above equation using the quotient rule of differentiation.
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{1 + \dfrac{{{x^2}\sin^{2}a}}{{{{\left( {1 – x\cos a} \right)}^2}}}}}\left[ {\dfrac{{\left( {1 – x\cos a} \right)\sin a – x\sin a\left( { - \cos a} \right)}}{{{{\left( {1 – x\cos a} \right)}^2}}}} \right]$
$ \Rightarrow $$\dfrac{{dy}}{{dx}} = \dfrac{{{{\left( {1 – x\cos a} \right)}^2}}}{{{{\left( {1 – x\cos a} \right)}^2} + {x^2}\sin^{2}a}}\left[ {\dfrac{{\sin a – x\sin a\cos a + x\sin a\cos a}}{{{{\left( {1 – x\cos a} \right)}^2}}}} \right]$
$ \Rightarrow $$\dfrac{{dy}}{{dx}} = \dfrac{{\sin a}}{{1 + {x^2}\cos^{2}a + 2x\cos a + {x^2}\sin^{2}a}}$
$ \Rightarrow $$\dfrac{{dy}}{{dx}} = \dfrac{{\sin a}}{{1 + {x^2}\left( {\cos^{2}a +\sin^{2}a} \right) + 2x\cos a}}$
$ \Rightarrow $$\dfrac{{dy}}{{dx}} = \dfrac{{\sin a}}{{1 + {x^2} + 2x\cos a}}$ [Since $\sin^{2}x + \cos^{2}x = 1$]
Now equate the above equation with the given differential equation.
We get, $A = \sin a$

Option ‘C’ is correct

Note: Whenever we get an equation that is a mix of $x$ and $y$, we need to rewrite the equation as y in terms of $x$. After that, we can find the derivative of the equation and compare the given derivative value with the calculated one to get the value of $A$.