Question

The solution of $\tan y\frac{dy}{dx}=\sin (x+y)+\sin (x-y)$ is

• A. $\sec y=2\cos x+c$
• B. sec $y=-2\cos x+c$
• C. $\tan y=-2\cos x+c$
• D. ${\sec }^{2}y=-2\cos x+c$

Answer 1

Answer: (B)

$\tan y\frac{dy}{dx}=\sin (x+y)+\sin (x-y)$
$\tan y\frac{dy}{dx}=2\cdot \sin \left(\frac{2x}{2}\right)\cdot \cos \left(\frac{2y}{2}\right)$
$\left[\because \sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cdot \cos \left(\frac{C-D}{2}\right)\right]$
$\Rightarrow \tan y\frac{dy}{dx}=2\sin x\cdot \cos y$
$\Rightarrow \frac{\sin y}{\cos y}\frac{dy}{dx}=2\sin x\cos y$
On integrating
$\Rightarrow \int \frac{\sin y}{{\cos }^{2}y}dy=\int 2\sin xdx$
$\left[\begin{array}{c}\text{ Let }t=\cos y\\ \frac{dt}{dy}=-\sin y\\ -dt=\sin ydy\end{array}\right]$
$\Rightarrow -\int \frac{dt}{t^2}=2(-\cos x)+c$
$\Rightarrow -\left(-\frac{1}{t}\right)=-2\cos x+c$
$\Rightarrow \frac{1}{\cos y}=-2\cos x+c$
$\Rightarrow \sec y=-2\cos x+c$