Question

The solution of the differential equation

$\cos y \log ( \sec x + \tan x ) d x = \cos x \log ( \sec y + \tan y ) d y$ is

• A. $\sec ^ { 2 } x + \sec ^ { 2 } y = c$
• B. $\sec x + \sec y = c$
• C. $\sec x - \sec y = c$
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

$\cos y \log ( \sec x + \tan x ) d x = \cos x \log ( \sec y + \tan y ) d y$

⇒ $\int \sec y \log ( \sec y + \tan y ) d y$

$= \int \sec x \log ( \sec x + \tan x ) d x$

Put $\log ( \sec x + \tan x ) = t$and$\log ( \sec y + \tan y ) = z$

$\frac { [ \log ( \sec x + \tan x ) ] ^ { 2 } } { 2 } = \frac { [ \log ( \sec y + \tan y ) ] ^ { 2 } } { 2 } + c$ .