Question

If $\sin y=x\sin (a+y)$ and $\frac{dy}{dx}=\frac{A}{1+x^2-2x\cos a}$, then the value of $A$ is

• A. $2$
• B. $\cos a$
• C. $\sin a$
• D. None of these

Answer 1

The correct option is C

$\sin a$

Explanation for the correct option:

Step 1: Rearrange the given function in terms of $y$,

Use the trigonometric formula $\sin (a+b)=\sin a\cos b+\cos a\sin b$

Given $\sin y=x\sin (a+y)$, then expanding and rearranging, we get

$\begin{array}{c}\sin y=x\sin (a+y)\\ \sin y=x(\sin a\cos y+\cos a\sin y)\\ \sin y=x\sin a\cos y+x\cos a\sin y\\ \sin y-x\cos a\sin y=x\sin a\cos y\\ \sin y(1-x\cos a)=x\sin a\cos y\\ \frac{\sin y}{\cos y}=\frac{x\sin a}{1-x\cos a}\\ \tan y=\frac{x\sin a}{1-x\cos a}\\ y={\tan }^{-1}\left(\frac{x\sin a}{1-x\cos a}\right)........\left(i\right)\end{array}$

Step 2: Differentiate equation (i) with respect to $x$,

Use the Chain rule states that the derivative of $f\left(g\right(x\left)\right)$ is $f\text{'}\left(g\right(x\left)\right)\cdot g\text{'}\left(x\right)$, we also know that ${\sin }^{2}a+{\cos }^{2}a=1$.

$\begin{array}{c}y={\tan }^{-1}\left(\frac{x\sin a}{1-x\cos a}\right)\\ \frac{dy}{dx}=\frac{1}{1+{\left(\frac{x\sin a}{1-x\cos a}\right))}^2}\times \frac{(1-x\cos a)\times \sin a-x\sin a(0-\cos a)}{{(1-x\cos a)}^2}\\ \frac{dy}{dx}=\frac{1}{1+\frac{x^2{\sin }^{2}a}{{(1-x\cos a)}^2}}\times \frac{\sin a-x\sin a\cos a+x\sin a\cos a}{{(1-x\cos a)}^2}\\ \frac{dy}{dx}=\frac{{(1-x\cos a)}^2}{{(1-x\cos a)}^2+x^2{\sin }^{2}a}\times \frac{\sin a}{{(1-x\cos a)}^2}\\ \frac{dy}{dx}=\frac{\sin a}{1-2x\cos a+x^2{\cos }^{2}a+x^2{\sin }^{2}a}\\ \frac{dy}{dx}=\frac{\sin a}{1-2x\cos a+x^2\left({\cos }^{2}a+{\sin }^{2}a\right)}\\ \frac{dy}{dx}=\frac{\sin a}{1-2x\cos a+x^2}......\left(ii\right)\end{array}$

Step 3: Compare the Right-hand side of the above derivative with the given derivative in the question,

$$\begin{gathered} \frac{dy}{dx}=\frac{A}{1+x^2-2x\cos a},\text{and} \\ \frac{dy}{dx}=\frac{\sin a}{1-2x\cos a+x^2} \end{gathered}$$

Therefore, we get that the value of $A$ is $\sin a$.

Hence, the correct option is (C).