Question

Solution of $\tan y.{\sec }^{2}xdx+\tan x.{\sec }^{2}ydy=0$ is:

• A. $\sec x\sec y=c$
• B. $\tan x.\tan y=c$
• C. $\sin x.\sin y=c$
• D. $\cos x.\cos y=c$

Answer 1

Answer: (B)

$\tan x.\tan y=c$

$tanyse{c}^{2}xdx+tanxse{c}^{2}ydy=0$

$\Rightarrow \frac{se{c}^{2}xdx}{tanx}+\frac{se{c}^{2}y}{tany}dy=0$

let $\begin{array}{cc}tanx=t;& tany=k\\ \Rightarrow se{c}^{2}xdx=dt& \Rightarrow se{c}^{2}ydy=k\end{array}$

$\Rightarrow \int \frac{dt}{t}+\int \frac{dk}{k}=c$

$\Rightarrow logkt=c$

$\Rightarrow kt=c$

$\Rightarrow tanxtany=c$