Question

prove tan A + tan B ÷ cot A + cot B = tan A × tan B

Answer 1

Hello Friend,


Here,

LHS
= (tan A + tan B) ÷ (cot A + cot B)

[ Now, cot θ = 1/tan θ ]

= (tan A + tan B) ÷ [(1/tan A) + (1/tan B)]
= (tan A + tan B) ÷ [ (tan A + tan B)/(tan A tan B)]

[Here, there is (tan A + tan B) both in numerator as well as denominator. So it gets cancelled]

= tan A tan B
= RHS

Hence proved.

Hope it helps.

Purva
@Purvaparmar1405
Brainly.in

Answer 2

Hello friend

Here's your answer

tan A + tan B / cot A + cot B

We know that cot A = 1 / tan A

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

= tan A + tan B / (tanA+tanB) /(tanA × tan B)

=(tan A + tan B) × (tan A×tan B) / (tan A + tan B)

By cancelling tan A + tan B

= tan A × tan B