proving...(tan^3x + sinxsecx - sinxcosx)/ (secx- cosx) = tanxsecx+sinx?

Question

(tan^3x + sinxsecx - sinxcosx)/ (secx- cosx) = tanxsecx+sinx

Answer 1

(tan^3x + sinxsecx - sinxcosx)/ (secx- cosx)
=(tan^2x*sinx + sinx - sinx*cos^2x)/( 1-cos^2x)
=sinx(tan^2x + (1 - cos^2x))/sin^2x
=(tan^2x + sin^2x)/sinx
=tanx*sinx/(cosx*sinx) + sin^2x/sinx
=tanx*secx + sinx

Answer 2

(tan^3x + sinxsecx - sinxcosx)/ (secx- cosx) = tanxsecx+sinx

LHS:
(tan^3x + sinxsecx - sinxcosx)/ (secx- cosx)
covert sec x into 1/cosx
=[tan^3x + sinx(1/cox x)-sinxcosx] / [(1/cosx)-cosx)]
=[tan^3x + (sinx/cosx)-sinxcosx] / [(1/cosx)-cosx)]
simplify sinx/cosx to tan x and combine the denominator to form 1 fraction
=[tan^3x + tanx -sinxcosx] / [(1-cos²x)/cosx]
covert 1-cos²x to sin²x
(you need to know the identity sin²x+cos²x=1)
=tan^3x + tanx -sinxcosx] / [sin²x / cosx]
covert sin²x/cosx to sinxtanx (cause sinx/cosx = tanx)
=tan^3x + tanx -sinxcosx] / (sinxtanx)
divide the terms in the numerator by tanx
=(tan²x+1-cos²x) / sinx
convert tan²x to sin²x/cos²x and 1-cos² to sin²x (using the same identity as shown above)
=[(sin²x/cos²x)+sin²x] / sinx
divide the numerator by sin x thoroughout
=(sinx/cos²x)+sinx
change sinx/cosx to tanx
=tanx(1/cosx)+sinx
convert 1/cosx to sec x
=tanxsecx+sinx
=RHS (proven)

Hope this helps!

Answer 3

First convert everything into sines and cosines: sin^3/cos^3+sin/cos-sincos)/(1/cos-cos/1)
This expression simplifies to: [sin^3/cos^3 +sincos^2/cos^3-sincos^4/cos^3]cos/sin^2 and this simplifies to: [sin/cos^2+cos^2/(sincos^2)-cos^4/(sincos^2)]. The first term, sin/cos^2 is simply tanxsecx; the next two terms, cos^2/(sincos^2)-cos^4/(sincos^2) equals 1/sin-cos^2/sin which equals [1-cos^2]/sin = sin^2/sin=sin....and that is what you wanted.