prove that: (tanA-secB)(cotA-cosB)=tanAcosB-cotAsecB

We have to prove that (tan A - sec B)(cot A - cos B) = tan
A *cos B - cot A * sec B.

We use the definitions: tan x =
sin x / cos x, cot x = 1/tan x and sec x = 1/ cos x

(tan A
- sec B)(cot A - cos B)

=> tan A*cot A - tan A*cos B
- sec B*cot A + sec B*cos B

=> 1 - tan A*cos B - sec
B*cot A + 1

As can be seen we cannot get the required
result using this.

Instead of (tan A - sec B)(cot A - cos
B), it should be (tan A - sec B)(cot A + cos B). In that
case:

(tan A - sec B)(cot A + cos
B)

=> tan A * cos B + tan A * cot A - sec B * cot A
- sec B*cos B

=> tan A * cos B + 1 - sec B * cot A -
1

=> tan A * cos B - cot A * sec
B

The accurate identity using the given
trigonometric functions is (tan A - sec B)(cot A + cos B) = tan A*cos B - cot A * sec
B