2007-03-10 10:14:06 · 5 answers · asked by ma.roddy 1 in Science & Mathematics ➔ Mathematics
You could express it as cotx + tanx if you wanted
Whether or not that is a simplification is a matter of point of view.
2007-03-10 10:27:11 · answer #1 · answered by gandalf 4 · 0⤊ 0⤋
Reduce the fractions to the same denominator
cosx/sinx + sinx/cosx = (cos^2x + sin^2x)/sinxcosx = 1/sinxcosx , since cos^2x + sin^x = 1
You can reduce further using sin2x = 2 sinxcosx in the donominatos to obtain finally
1/sinxcosx = 2/sin2x
2007-03-10 10:20:23 · answer #2 · answered by physicist 4 · 0⤊ 0⤋
Cosx/Sinx +Sinx/Cosx
=Cosx(cosx) + Sinx(sinx)/Sinxcosx
=Cos^ x +Sin^x / Sinxcosx
=1/SinxCosx
=Cosecx .Secx
2007-03-10 16:21:30 · answer #3 · answered by ash 2 · 0⤊ 0⤋
Find the common denominator.
(cosx/sinx)(cosx/cosx)
+(sinx/cosx)(sinx/sinx)
(cos^2x+sin^2x)/(sinxcosx)
The numerator simplifies to 1
according to the pythagorean identity.
The denominator simplifies to .5sin(2x).
1/(.5sin(2x))
=2csc(2x)
2007-03-10 10:20:30 · answer #4 · answered by Tim 2 · 0⤊ 0⤋
= (cos²x + sin²x) / (sinx.cosx)
= 1 / sin x. cos x
= 2 / 2.sinx.cosx
= 2 / sin 2x
2007-03-10 10:21:31 · answer #5 · answered by Como 7 · 0⤊ 0⤋