Question

Which one of the following is the correct solution of $\left(\frac{dy}{dx}\right)\tan y=\sin \left(x+y\right)+\sin \left(x-y\right)$?

• A. $secx=C-2secy$
• B. $secy=C+2\cos y$
• C. $secy=C-2\cos x$
• D. $secx=C-2\cos y$

Answer 1

The correct option is C

$secy=C-2\cos x$

The explanation for the correct option:

Step 1: Rewrite the given equation

$\left(\frac{dy}{dx}\right)\tan y=\sin \left(x+y\right)+\sin \left(x-y\right)$

$\Rightarrow \left(\frac{dy}{dx}\right)\tan y=\sin x\cos y+\cos x\sin y+\sin x\cos y-\cos x\sin y$

$\Rightarrow$$\left(\frac{dy}{dx}\right)\tan y=2\sin x\cos y$

$\Rightarrow$$\frac{\tan y}{\cos y}dy=2\sin xdx$

Step 2: Integrating on both sides

We know that $\int \tan xsecxdx=secx+C$ and $\int K\sin xdx=-k\cos x+C$ where $C$ is an integration constant and $k$ is a constant.

$\int \tan ysecydy=\int 2\sin xdx$

$\Rightarrow$ $secy=-2\cos x+C$ [$C$ is an integration constant]

$\Rightarrow$ $secy=C-2\cos x$

Hence, the correct option is C