Question

under root 1+sinA upon 1-sinA=tanA+secA​

Answer 1

hey mate

here is ur answer

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Required to prove:

$\bf\sqrt {\frac {1+sinA}{1-sinA}}=tanA+secA$

Method :

$\bf consider \:LHS \sqrt {\frac {1+sinA}{1-sinA}×\frac{1+sinA}{1-sinA}} \\\ = \frac{(1+sinA)^2}{1^2-sin^2A}$

$\bf \underline {cos^2A=1-sin^2A}$

$\bf then\\=\frac {(1+sinA)^2}{(cosA)^2}$

$\underline{\frac {1}{cosA}=secA}$

$\underline{\frac {sinA}{cosA}=tanA}$

$\bf\underline {answer \: is \: secA+tanA =RHS}$

LHS=RHS

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hope the answer helps