# 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...

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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

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