Question

√[1+sinA/1-sinA] = sec A + tan A

Answer 1

√1+sin a/1-sina

we will first rationalise it

√1+sin a× 1+sin a/1-sin a × 1+sin a

√(1+sin a )²/1²-sin²a

square and root will be cancelled in the nr.

1+sina /√cos²a                                              .....        (sin²a+cos²a=1)identity

1+sin a /cos a = lhs

now rhs

sec a + tan a      (tan a =sina /cosa)

1/cos a+ sin a/cos a

1+sin a/ cos a =rhs

hence lhs =rhs

∴proved

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