Question

Prove that: √1+sinA/1-sinA=secA+tanA​

Answer 1

Step-by-step explanation:

√1-sinA/1+sinA=secA-tanA

L.H.S

√1-sinA/1+sinA

multiple by 1-SinA

√1-sinA/1+sinA×1-SinA/1-SinA

√(1-SinA)^2/√1-(Sin^2A)

1-SinA/√Cos^2

1-SinA/CosA

1/CosA-SinA/CosA

SecA-TanA

Hence Proved

Answer 2

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