A) \[\tan x.\cot y=C\]
B) \[\cot x.tany=C\]
C) \[\tan x.tany=C\]
D) \[\sin x.\cos y=C\]
Correct Answer: C
Solution :
Given, \[{{\sec }^{2}}y.\tan \,y\,dx+{{\sec }^{2}}y.\tan x\,dy=0\] On separating the variables, we get \[\Rightarrow \]\[{{\sec }^{2}}x.\tan ydx=-{{\sec }^{2}}y.\tan x\,dy\] \[\Rightarrow \]\[\frac{{{\sec }^{2}}x}{\tan x}dx=-\frac{{{\sec }^{2}}y}{\tan y}dy\] On integrating both the sides, we get \[\int\limits_{{}}^{{}}{\frac{{{\sec }^{2}}x}{\tan x}dx}=-\int\limits_{{}}^{{}}{\frac{{{\sec }^{2}}x}{\tan y}}dy\] Let \[\tan x=u\] \[\Rightarrow \] \[{{\sec }^{2}}x=\frac{du}{dx}\] \[\Rightarrow \] \[dx=\frac{du}{{{\sec }^{2}}y}\]and \[\tan y=v\] \[\Rightarrow \] \[{{\sec }^{2}}y=\frac{dv}{dy}\] \[\Rightarrow \] \[dy=\frac{du}{{{\sec }^{2}}y}\] \[\int\limits_{{}}^{{}}{\frac{{{\sec }^{2}}x}{u}.\frac{du}{{{\sec }^{2}}x}}=-\int\limits_{{}}^{{}}{\frac{{{\sec }^{2}}y}{v}.\frac{dv}{{{\sec }^{2}}y}}\] \[\Rightarrow \] \[\int_{{}}^{{}}{\frac{du}{u}=-\int_{{}}^{{}}{\frac{dv}{v}}}\] \[\Rightarrow \]\[\log |x|=-\log |v|+\log |c|\] \[\Rightarrow \]\[\log |\tan x|=-\log |\tan \,y|+\log |c|\] \[\Rightarrow \]\[\log |\tan \,x.\tan \,y|=\log |c|\] \[(\because \,\log \,m+\log \,n=\log \,mn)\] \[\Rightarrow \] \[(\because \,\log \,m=\log \,n\Rightarrow m=n)\] which is the required general solution;
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