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
