Asked by Jay
We are doing trig identities in school. I need help with these five:
1.1+sinx/1-sinx=cscx+1/cscx-1
2.tanx+sinx/1+cosx=tanx
3.sec^2x/sin^2x=1/sin^2x+1/cos^2x
4.tan^2x/1+tan^2x=1-cos^2x
5.sinx-sin^3x/cos^3x=tanx
1.1+sinx/1-sinx=cscx+1/cscx-1
2.tanx+sinx/1+cosx=tanx
3.sec^2x/sin^2x=1/sin^2x+1/cos^2x
4.tan^2x/1+tan^2x=1-cos^2x
5.sinx-sin^3x/cos^3x=tanx
Answers
Answered by
Reiny
My method
1. start with the more complicate looking side
2. change everything to sines and cosines, unless I recognize an obvious identity
I will do #2 for you
LS = (tanx + sinx)/(1 + cosx)
= (sinx/cosx + sinx)/(1+cosx)
= [( sinx + sinxcosx)/cosx ]/(1+cosx)
= [sinx(1+cosx)/cosx]/(1+cosx)
= sinx/cosx
= tanx
= RS
by "obvious identity" I notice in #4 that we have the expression
1 + tan^2x
from the Pythagorean identities
1 + tan^2x = sec^2x
so #4
LS = tan^2x/(1+tan^2)
= tan^2x/sec^2x
= tan^2x cos^2x
= sin^2x/cos^2x (cos^2x) = sin^2x
= 1 - cos^2x
= RS
1. start with the more complicate looking side
2. change everything to sines and cosines, unless I recognize an obvious identity
I will do #2 for you
LS = (tanx + sinx)/(1 + cosx)
= (sinx/cosx + sinx)/(1+cosx)
= [( sinx + sinxcosx)/cosx ]/(1+cosx)
= [sinx(1+cosx)/cosx]/(1+cosx)
= sinx/cosx
= tanx
= RS
by "obvious identity" I notice in #4 that we have the expression
1 + tan^2x
from the Pythagorean identities
1 + tan^2x = sec^2x
so #4
LS = tan^2x/(1+tan^2)
= tan^2x/sec^2x
= tan^2x cos^2x
= sin^2x/cos^2x (cos^2x) = sin^2x
= 1 - cos^2x
= RS
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