Question

prove that root 1+sinA/1-sinA=secA+tanA
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Answer 1

Here's the answer

Hope it helps

Answer 2

Step-by-step explanation:

let us consider LHS

=√1+sinA/1-sinA

by rationalising the denominator

=√1+sinA/1-sinA(1+sinA)/1+sinA

=√(1+sinA)²/(1)²-(sinA)²

=√(1+sinA)²/√1-sin²A

root and square will be cancelled

=1+sinA/√1-sin²A

sin²a+cos²a= 1

cos²a=1-sin²a

cosa=√1-sin²a

=1+sinA/cosA

=1/cosA+sinA/cosA

=secA+tanA (RHS)

since, LHS=RHS

hence proved