Question

The solution of the differential equation

$\frac { d y } { d x } + \frac { 1 + \cos 2 y } { 1 - \cos 2 x } = 0$

• A. $\tan y + \cot x = c$
• B. $\tan y \cot x = c$
• C. $\tan y - \cot x = c$
• D. None of these

Answer 1

Answer: (C)

$\tan y - \cot x = c$

Answer 2

$\frac { d y } { d x } = - \frac { 1 + \cos 2 y } { 1 - \cos 2 x }$ ⇒ $\frac { d y } { d x } = - \frac { 2 \cos ^ { 2 } y } { 2 \sin ^ { 2 } x }$

⇒ $\sec ^ { 2 } y d y = - \operatorname { cosec } ^ { 2 } x d x$

On integrating both sides, we get

$\tan y = \cot x + c$ ⇒ $\tan y - \cot x = c$.