A) \[y\text{ }sec\text{ }x=tan\text{ }x+c\]
B) \[y\text{ }tan\text{ }x=sec\text{ }x+c\]
C) \[tan\text{ }x=y\text{ }tan\text{ }x+c\]
D) \[x\text{ }sec\text{ }x=tan\text{ }y+c\]
E) \[x\text{ }tan\text{ }x=y\text{ }tan\text{ }x+c\]
Correct Answer: A
Solution :
Given equation is \[\frac{dy}{dx}+y\tan x=\sec x\] Here,\[P=tan\text{ }x\]and\[Q=sec\text{ }x\] \[\therefore \] \[IF={{e}^{\int{P\,dx}}}={{e}^{\int{\tan x\,dx}}}\] \[={{e}^{\log x\sec x}}=\sec x\] \[\therefore \]Solution is \[y.\sec x=\int{{{\sec }^{2}}x}\,dx+c\] \[\Rightarrow \] \[y.\sec x=\tan x+c\]
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