Question

(sinx/1-cosx)-(sinxcosx/1+cosx)=cscx(1+cos^2x)

Answer 1

RHS = 1+ cos x
LHS = (sinx ( 1+ cos x ) - (sinx*cos^2x))/sin^2 x
simplify you get (cos x - sin^3x)/sin^2x
cos x - sin^3 x = sin^2 x
What are we to do exactly? And i did that mentally so it might be wrong just check it please :P

Answer 2

Sorry we are supposed to verify that each trigonometric equation is an identity.

Answer 3

Oh... doing this mentally is so much harder than i thought =.=
See i have a simple trick... What you do is simplify every trigonometric notion into simplest form, and keep using identities. here the only one which is of any use is sin^2 + cos^2 = 1
So you keep using that and if you cant ask me ill try again :)

Answer 4

Ok thanks!

Answer 5

Numerator of the left side of the identity:
sin x(1 + cos x)- sin x*cos x(1 - cos x) =
= sin x + sin x*cos x - sin x*cos x + sin x cos^2 x = sin x(1+ cos ^2x)
Denominator: (1 - cos x)(1 + cos x) = 1 - cos^2 x = sin^2 x
Left side after simplification: (1 + cos^2 x)/sin x = csc x(1 + cos ^2 x)
That proves the identity.

Answer 6

^ Perfect :)