I am sure you meant :
tan2x+sec2x = (cosx+sinx)/(cosx-sinx)
LS
= sin2x/cos2x + 1/cos2x
= (sin2x +1)/cos2x
= (2sinxcosx + sin^2x + cos^2x)/(cos^2x - sin^2x)
= (sinx + cosx)^2/[(cosx-sinx)(cosx+sinx)]
= (sinx + cosx)/(cosx - sinx)
= RS
Q.E.D.
Prove:
tan2x+sec2x = cosx+sinx/ cosx-sinx
3 answers
This time start from the right-hand side by taking advantage of the term cos(x)-sin(x):
(cos(x)+sin(x))/(cos(x)-sin(x))
multiply top and bottom by cos(x)+sin(x)
(cos(x)+sin(x))^sup2;/(cos²(x)-sin²(x)
=(cos²(x)+sin²(x)+2sin(x)cos(x))/(cos²(x)-sin²(x))
=(1+sin(2x))/cos(2x)
=sec(2x)+tan(2x)
(cos(x)+sin(x))/(cos(x)-sin(x))
multiply top and bottom by cos(x)+sin(x)
(cos(x)+sin(x))^sup2;/(cos²(x)-sin²(x)
=(cos²(x)+sin²(x)+2sin(x)cos(x))/(cos²(x)-sin²(x))
=(1+sin(2x))/cos(2x)
=sec(2x)+tan(2x)
Thank you so much Reiny and Mathmate. Both of you guys gave me great ways to solving this problem.
2 answers