SOLUTION: prove Sin(90-θ)/sec(90- θ)*tan(90- θ)/cos θ=cos θ

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Question 212096: prove
Sin(90-θ)/sec(90- θ)*tan(90- θ)/cos θ=cos θ
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Note: In order to use the Algebra.com's software more effectively, I am going to use "x" instead of "θ".
%28sin%2890-x%29%2Fsec%2890-+x%29%29%2A%28tan%2890-+x%29%2Fcos%28x%29%29+=+cos%28x%29

Often the easiest way to prove identities is to use basic identities to convert all other trig functions into sin and/or cos. So we will start by replacing sec and tan using sec%28x%29+=+1%2Fcos%28x%29 and tan%28x%29+=+sin%28x%29%2Fcos%28x%29:


To simplify this compound fraction, all we need to do is multiply the top and bottom of the "big" fraction by cos(90- x):

Now we can cancel the cos(90-x)'s:

leaving
%28sin%2890-x%29%2Asin%2890-x%29%29%2Fcos%28x%29+=+cos%28x%29
Since sin(90-x) = cos(x) we can substitute:
%28cos%28x%29%2Acos%28x%29%29%2Fcos%28x%29+=+cos%28x%29
We can cancel a cos(x):
%28cross%28cos%28x%29%29%2Acos%28x%29%29%2Fcross%28cos%28x%29%29+=+cos%28x%29
leaving
cos%28x%29+=+cos%28x%29

Note: This is just one way to prove this identity. There are many others.