Question

prove that √(1+sinA/1-sinA) + √(1-sinA/1+sinA) = 2secA.​

Answer 1

Answer:

Given:

The equation is,

√(1+sinA/1-sinA)+√(1-sinA/1+sinA)=2secA

To prove:

LHS=RHS

Solution:

First take LHS,

√(1+sinA/1-sinA)+√(1-sinA/1+sinA)

on rationalizing,

=√(1+sinA/1-sinA)×1+sinA/1+sinA)+√(1-sinA/1+sinA)×1-sinA/1-sinA)

=√(1+sinA)²/1-sin²A)+√(1-sin²A)/(1+sinA)²

=1+sinA/cosA+cosA/1+sinA

=(1+sinA)²+cos²A/cosA(1+sinA)

=1+sin²A+cos²A+2sinA/cosA (1+sinA)

=2+2sinA/cosA (1+sinA)

=2(1+sinA)/cosA(1+sinA)

=2/CosA

=2SecA

Hence proved.....

Step-by-step explanation:

Hope it helps you.......