Question

prove: (seca-secb)/(tana+tanb)=(tan-tanb)/(seca+secb)

Answer 1

Step-by-step explanation:

LHS

(secA - secB) / (tanA + tanB)

by rationalising

(secA - secB)(secA + secB) / (secA + secB) × (tanA - tanB) / (tanA - tanB)(tanA + tanB)

(sec^2A - sec^2B) / (secA + secB) × (tanA - tanB) / (tan^2A - tan^2B)

{1 + tan^2A - (1 + tan^2B)} / (secA + secB) × (tanA - tanB) / (tan^2A - tan^2B)

(tan^2A - tan^2B) / (secA + secB) ×(tanA - tanB) / (tan^2A - tan^2B)

(tanA - tanB) / (secA - secB)