# Prove that : (tanA+cotB)(cotA-tanB)=cotAcotB-tanAtanB

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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)

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

tan x = 1/ cot x or cot x * tan x = 1

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

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

**This proves that (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

We'll manipulate the left side of trigonometric identity.

We'll substitute cot B = 1/tan B and cot A = 1/tan A

We'll re-write the left side:

(tanA+cotB)(cotA-tanB) = (tan A + 1/tan B)(1/tanA - tan B)

We'll remove the brackets using FOIL method:

(tan A + 1/tan B)(1/tanA - tan B) = tanA/tanA - tanA*tanB + 1/tanA*tanB - tanB/tanB

We'll simplify, we'll eliminate like termsĀ and we'll get:

(tan A + 1/tan B)(1/tanA - tan B) = 1/tanA*tanB - tanA*tanB

But 1/tanA*tanB = cotA*cotB

**(tan A + 1/tan B)(1/tanA - tan B) = cotA*cotBĀ - tanA*tanB = RHS**