How do you simplify #sin(a - b)/sina x cosb = 1 - cota x tanb#?

1 Answer
Lotusbluete
Nov 27, 2015

I'm taking a wild guess here that with #x#, you actually meant "x", the multiplication sign.

I'm also taking a second wild guess that "#cos b#" on your left side should have been a part of the denominator...

So, I think what you would like to prove is

#sin(a-b) / (sin a * cos b ) = 1 - cot a * tan b #

To prove this, let's use the following:

  1. #cot x = cos x / sin x #

  2. #tan x = sin x / cos x #

  3. #sin (x - y ) = sin x cos y - cos x sin y #

Now you can prove the identity as follows:

#sin(a-b) / (sin a * cos b ) = (sin a cos b - cos a sin b) / (sin a cos b) #

# color(white)(xxxxxxxx) = (sin a cos b) / (sin a cos b) - (cos a sin b) / (sin a cos b) #

# color(white)(xxxxxxxx) = 1 - (cos a * sin b) / (sin a * cos b) #

# color(white)(xxxxxxxx) = 1 - cos a / sin a * sin b / cos b #

# color(white)(xxxxxxxx) = 1 - cot a * tan b #

Hope that this helped.

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