trigonometry- identities?

Question

(1 + sin2x) / (cos2x) = (cosx + sinx) / (cosx - sinx)

(1 + sin2x) / (cos2x) = (cosx + sinx) / (cosx - sinx)

please prove that it is equal through application of general formulas...
you can either change the left or right side to verify the equation

Answer 1

Prove the identity.

[1 + (sin 2x)] / (cos 2x) = (cosx + sinx) / (cosx - sinx)

Right Hand Side = (cosx + sinx) / (cosx - sinx)

= (cosx + sinx)² / [(cosx - sinx)(cosx + sinx)]

= [cos²x + 2(cosx)(sinx) + sin²x] / (cos²x - sin²x)

= [(cos²x + sin²x) + 2(cosx)(sinx)] / (cos 2x)

= [1 + (sin 2x)] / (cos 2x) = Left Hand Side

Answer 2

sin2x = sin(x + x) = sinx*cosx + cosx*sinx = 2*sinx*cosx

also,

cos2x = cos(x + x) = cosx*cosx – sinx*sinx =
= (cosx)^2 – (sinx)^2 = (cosx + sinx)*(cosx – sinx)

and,

1 = (sinx)^2 + (cosx)^2

substitute the above on the left hand-side of the equation:

[(sinx)^2 + (cosx)^2 + 2*sinx*cosx]/[ (cosx + sinx)*(cosx – sinx)]

you know that:

(sinx)^2 + (cosx)^2 + 2*sinx*cosx = (sinx + cosx)^2

plug this into the numerator:

(sinx + cosx)^2/[(cosx + sinx)*(cosx – sinx)]

simplify:

(sinx + cosx)/(cosx – sinx)

Q.E.D.

Answer 3

(1 + sin2x)/(cos2x)
=[ sin^2(x) + cos^2(x) + 2sinxcos]/cos2x
= [(cosx + sinx)^2]/[cos^2(x) -- sin^2(x)]
= [cosx + sinx]/[cosx -- sinx] as in denominator
cos^2(x) -- sin^2(x)=(cosx+sinx)(cosx--sinx)

Answer 4

tel me what to do exactly then i will try to ans.