A) \[{{\sec }^{2}}x+{{\sec }^{2}}y=c\]
B) \[\sec x+\sec y=c\]
C) \[\sec x-\sec y=c\]
D) None of these
Correct Answer: D
Solution :
\[\cos y\log (\sec \,\,x+\tan x)dx=\cos x\log (\sec y+\tan y)dy\] Þ \[\int_{{}}^{{}}{\sec y\log (\sec y+\tan y)dy}\] \[=\int_{{}}^{{}}{\sec x\log (\sec x+\tan x)dx}\] Put \[\log (\sec x+\tan x)=t\]and\[\log (\sec y+\tan y)=z\] \[\frac{{{[\log (\sec x+\tan x)]}^{2}}}{2}=\frac{{{[\log (\sec y+\tan y)]}^{2}}}{2}+c\].
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