# prove the following: (tan A - sec B) (cot A + cos B) = tan A cos B - cot A sec B

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### 2 Answers

We have to prove that: (tan A - sec B) (cot A + cos B) = tan A cos B - cot A sec B

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

open the brackets and multiply the terms:

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

use tan x * cot x = 1 and sec x * cos x = 1

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

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

**This proves that (tan A - sec B) (cot A + cos B) = tan A cos B - cot A sec B**

Q: Prove : (tan A - sec B) (cot A + cos B) = tan A cos B - cot A sec B

A: L:H:S = (tan A - sec B) (cot A + cos B)

= tanA.cotA + tanA.cosB - cotA.secB - cosB.secB

we know : tanA.cot A = 1, cosB.secB = 1

= 1 +tanA.cosB - cotA.secB -1

= tanA.cosB - cotA.secB

L:H:S = R:H:S