If dy/dx = tan(x), then y=
a) 1/2 (tan x)^2 + c
b) (sec x)^2 + c
c) ln|sec x| + c
d) ln|cos x| + c
e) sec x tan x + c

Dy/dx= tanx, can be answered directly using the derivatives of trigonometric functions but this is how the answer is derived
         =(sinx/cosx) basic trigonometric function
         = [cosx cox+sinxsinx]/cos^2x
         =[cos^2x+sin^2x]/cos^2x
cos^2+sin^2x = 1 ; fundamental trigonometric identities
         = 1/cos^2x; reciprocal relations
          = sec^2x+C
The answer is letter B.sec^2x+C

The answer to this question is:

B) (sec x)^2 + C

I’m sorry i can’t u