prove cos(x+y) / cos(x-y) = (1-tan x tan y)/(1+tan x tan y)

LS = (cosxcosy - sinxsiny)/(cosxcosy + sinxsiny)

divide the numerator and the denominator by cosxcosy and see how it comes apart into the RS.

To prove the equation cos(x+y) / cos(x-y) = (1-tan x tan y)/(1+tan x tan y), we'll start with the left side and manipulate it algebraically to arrive at the right side.

Let's begin:

1. Start with the left side of the equation:
cos(x+y) / cos(x-y)

2. Using the cosine sum and difference identity, we can write cos(x+y) and cos(x-y) in terms of cosines of x and y. The identity is:
cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
cos(a-b) = cos(a)cos(b) + sin(a)sin(b)

Applying these identities to the equation, we get:
(cos(x)cos(y) - sin(x)sin(y)) / (cos(x)cos(y) + sin(x)sin(y))

3. Now, we need to rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator, which is (cos(x)cos(y) - sin(x)sin(y)):
[(cos(x)cos(y) - sin(x)sin(y)) * (cos(x)cos(y) - sin(x)sin(y))] / [(cos(x)cos(y) + sin(x)sin(y)) * (cos(x)cos(y) - sin(x)sin(y))]

Simplifying the numerator and denominator separately:
[cos^2(x)cos^2(y) - 2cos(x)cos(y)sin(x)sin(y) + sin^2(x)sin^2(y)] / [cos^2(x)cos^2(y) - sin^2(x)sin^2(y)]

4. Notice that cos^2(x)cos^2(y) and sin^2(x)sin^2(y) are perfect squares and can be expressed using the Pythagorean identity:
cos^2(x) = 1 - sin^2(x)
cos^2(y) = 1 - sin^2(y)

Taking these substitutions into account, the equation becomes:
[(1 - sin^2(x))(1 - sin^2(y)) - 2cos(x)cos(y)sin(x)sin(y)] / [(1 - sin^2(x))(1 - sin^2(y)) - sin^2(x)sin^2(y)]

5. Expanding the numerator:
(1 - sin^2(x) - sin^2(y) + sin^2(x)sin^2(y)) - 2cos(x)cos(y)sin(x)sin(y)
= 1 - sin^2(x) - sin^2(y) + sin^2(x)sin^2(y) - 2cos(x)cos(y)sin(x)sin(y)

6. Simplifying the denominator:
(1 - sin^2(x))(1 - sin^2(y)) - sin^2(x)sin^2(y)
= (1 - sin^2(x) - sin^2(y) + sin^2(x)sin^2(y)) - sin^2(x)sin^2(y)

7. Notice that the numerator and denominator have the same terms, so we can cancel them out:
(1 - sin^2(x) - sin^2(y) + sin^2(x)sin^2(y)) - 2cos(x)cos(y)sin(x)sin(y) / (1 - sin^2(x) - sin^2(y) + sin^2(x)sin^2(y)) - sin^2(x)sin^2(y)
= 1 - 2cos(x)cos(y)sin(x)sin(y)

8. Now, let's simplify the right side of the equation:
(1 - tan(x)tan(y))/(1 + tan(x)tan(y))

9. Recall that tan(x) = sin(x)/cos(x) and substitute into the equation:
(1 - (sin(x)/cos(x))(sin(y)/cos(y))) / (1 + (sin(x)/cos(x))(sin(y)/cos(y)))

10. Simplify both the numerator and denominator separately:
[(cos(x)cos(y) - sin(x)sin(y)) / (cos(x)cos(y))] / [(cos(x)cos(y) + sin(x)sin(y)) / (cos(x)cos(y))]

11. Invert the denominator and multiply:
[(cos(x)cos(y) - sin(x)sin(y))/ (cos(x)cos(y))] * [(cos(x)cos(y)) / (cos(x)cos(y) + sin(x)sin(y))]

12. Notice that the terms cancel out:
= (cos(x)cos(y) - sin(x)sin(y)) / (cos(x)cos(y) + sin(x)sin(y))

13. This is the same expression as the left side of the equation, which verifies the equation:
(cos(x+y) / cos(x-y)) = (1 - tan(x)tan(y)) / (1 + tan(x)tan(y))

Therefore, we have proven that the equation cos(x+y) / cos(x-y) = (1-tan x tan y)/(1+tan x tan y) is true.