Prove that sinA(1+tanA)+cosA(1+cotA) . . .

Question : Prove that sinA(1+tanA)+cosA(1+cotA) = secA+cosecA

Doubt by Afifa

Solution : OR

sinA(1+tanA)+cosA(1+cotA) = secA+cosecA

LHS
sinA(1+tanA)+cosA(1+cotA)
=sinA(1+sinA/cosA) + cosA(1+cosA/sinA)

[∵tanA=sinA/cosA, cotA=cosA/sinA]
=sinA[(cosA+sinA)]/cosA+cosA[(sinA+cosA)]/sinA
=sinA[(sinA+cosA)]/cosA+cosA[(sinA+cosA)]/sinA

=sinA+cosA[sinA/cosA+cosA/sinA]
=sinA+cosA[(sin²A+cos²A)/sinAcosA]
=sinA+cosA[1/sinAcosA]
[∵sin²A+cos²A=1]

=(sinA+cosA)/sinAcosA
=sinA/sinAcosA + cosA/sinAcosA
= 1/cosA + 1/sinA

= secA+cosecA
[∵1/cosA=secA and 1/sinA=cosecA]
= RHS

LHS=RHS
Hence Proved.

Class 10