Question

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

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