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

*print*Print*list*Cite

### 2 Answers

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

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

open the brackets and multiply the terms.

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

Use the relation tan x * cot x = 1

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

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

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

**This proves the identity (tan A + cot B) (cot A - tan B) = cot A cot B - tan A tan B**

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

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

= tanA.cotA - tanA.tanB + cotA.cotB -tanB.cotB

we know that : tanx.cotx =1

= 1-tanA.tanB+cotA.cotB -1

= cotA.cotB - tanA.tanB

Hence L:H:S = R:H:S