prove that √1+sinA/1-sinA=secA+tanA
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heya
it's too easy
√1+sinA/1-sinA×√1+sinA/1+sinA( multiply by √1+sinA on both side)
=√(1+sinA)^2/√1-sin^2A
=(√1+sinA)^2/√Cos^2A
=1+sinA/cosA
=secA+tanA
hope it help you
@rajukumar.
it's too easy
√1+sinA/1-sinA×√1+sinA/1+sinA( multiply by √1+sinA on both side)
=√(1+sinA)^2/√1-sin^2A
=(√1+sinA)^2/√Cos^2A
=1+sinA/cosA
=secA+tanA
hope it help you
@rajukumar.
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