proving/simplify trig

detectivea0149

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May 22, 2015
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secx+cscx-cosx-sinx=sinxtanx+cosxcotx

This is as far as I got
(1/cosx)+(1/sinx)-cosx-sinx=sinx(sinx/cosx)+cosx(cosx/sinx) I turned them into sin and cos
(sinx/sinxcosx)+(cosx/sinxcosx)-cosx-sinx = (sin2x/cosx)+(cos2x/sinx) I multiplied the second equation and I tried to get similar denominators in the first equation
I'm lost at what to do next.
 
secx+cscx-cosx-sinx=sinxtanx+cosxcotx

This is as far as I got
(1/cosx)+(1/sinx)-cosx-sinx=sinx(sinx/cosx)+cosx(cosx/sinx) I turned them into sin and cos
(sinx/sinxcosx)+(cosx/sinxcosx)-cosx-sinx = (sin2x/cosx)+(cos2x/sinx) I multiplied the second equation and I tried to get similar denominators in the first equation
I'm lost at what to do next.

L.H.S = \(\displaystyle \displaystyle{\frac{1}{cos(x)} + \frac{1}{sin(x)} - [sin(x) + cos(x)]}\)

= \(\displaystyle \displaystyle{\frac{cos(x) + sin(x)}{sin(x) * cos(x)} - [sin(x) + cos(x)]}\)

= \(\displaystyle \displaystyle{\frac{[cos(x) + sin(x)] - [sin(x) * cos(x)] * [sin(x) + cos(x)]}{sin(x) * cos(x)}}\)

= \(\displaystyle \displaystyle{\frac{[cos(x) + sin(x)] * [1 - sin(x) * cos(x)]}{sin(x) * cos(x)}}\)

= \(\displaystyle \displaystyle{\frac{[cos(x) + sin(x)] * [cos^2(x) + sin^2(x) - sin(x) * cos(x)]}{sin(x) * cos(x)}}\)

recall a3 + b3 = (a+b)(a2 -ab + b2)

= \(\displaystyle \displaystyle{\frac{cos^3(x) + sin^3(x)}{sin(x) * cos(x)}}\)

continue.....
 
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