tanX sinX + cosX = secX

Question

trigonometric identities

Answer 1

tanxsinx + cosx = secx , ,(sin^2x)/cosx + cosx/cosx = 1/cosx, ,sin^2x + cosx = 1 , ,(1 - cos2x)/2 + cosx = 1 , ,1 - cos2x + 2cosx = 2 , ,1 - (2cos^2x - 1) + 2cosx = 2, ,2cos^2x + 2cosx -2 =0, ,(2cosx - 1)^2 = 0, ,cosx = 0.5, ,basic angle = 60, ,x = 60, 300 degrees

Answer 2

"Solution: Given tan X sin X + cos X = sec X
LHS tan X sin X+cos X because (tan x=sin x/cos x)
= sin² X/cos X+cos X
=(sin² X+cos² X)/cos X because (sin² x+cos² x=1)
=1/cos X
=sec X
hence RHS
there for LHS=RHS"

Derivative of Secx Tanx - http://math.tutorcircle.com/calculus/derivative-of-secx-tanx.html