Asked by Haley
verify that the equation is an identity:
(1-cosX)/(1+cosX) = (cscX-cotX)^2
(1-cosX)/(1+cosX) = (cscX-cotX)^2
Answers
Answered by
Reiny
LS = (1- cosx)/(1 + cosx) [(1-cox)/(1-cosx)]
= (1-cosx)^2/(1-cos^2x)
RS = (1/sinx - cosx/sinx)^2
= ((1-cosx)/sinx)^2
= (1-cosx)^2 /(sin^2x)
= (1-cosx)^2/(1-cos^2)
= LS
= (1-cosx)^2/(1-cos^2x)
RS = (1/sinx - cosx/sinx)^2
= ((1-cosx)/sinx)^2
= (1-cosx)^2 /(sin^2x)
= (1-cosx)^2/(1-cos^2)
= LS
Answered by
Haley
thanks. How would you know how to do similar problems? Where should i start when verifying trig equations?
Answered by
Reiny
Usually start with the complicated looking side and try to reduce to the simpler side.
In this case I worked on both sides.
Also I often change everything to sines and cosines, unless I recognize a popular identity.
How did I know where to start??
Let's say doing this for 35 years and doing thousands of these could be the reason, lol
In this case I worked on both sides.
Also I often change everything to sines and cosines, unless I recognize a popular identity.
How did I know where to start??
Let's say doing this for 35 years and doing thousands of these could be the reason, lol
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