Question

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Answer 1

Solution →

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$\sf{\sqrt{\dfrac{1+sinA}{1-sinA} } }\\\\\\\sf{=\sqrt{\dfrac{(1+sinA)(1+sinA)}{(1-sinA)(1+sinA)} } }\\\\\\\sf{=\sqrt{\dfrac{(1+sinA)^2}{(1)^2-(sinA)^2} } }\\\\\\\sf{=\sqrt{\dfrac{(1+sinA)^2}{1-sin^2A} } }\\\\\\\sf{=\sqrt{\dfrac{(1+sinA)^2}{cos^2A} } }\\\\\\\sf{=\sqrt{\dfrac{(1+sinA)^2}{(cosA)^2} } }\\\\\\\sf{=\sqrt{\bigg(\dfrac{1+sinA}{cosA} \bigg)^2 } }\\\\\\\sf{=\dfrac{1+sinA}{cosA} } \\\\\\\sf{=\dfrac{1}{cosA}+\dfrac{sinA}{cosA} }\\\\\\\sf{=secA+tanA}$

Therefore, the answer is (B) sec A + tan A.

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Identities used :--

1 - sin²A = cos²A

1/cos A = sec A

sin A/cos A = tan A

Answer 2

Solution →

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$\sf{\sqrt{\dfrac{1+sinA}{1-sinA} } }\\\\\\\sf{=\sqrt{\dfrac{(1+sinA)(1+sinA)}{(1-sinA)(1+sinA)} } }\\\\\\\sf{=\sqrt{\dfrac{(1+sinA)^2}{(1)^2-(sinA)^2} } }\\\\\\\sf{=\sqrt{\dfrac{(1+sinA)^2}{1-sin^2A} } }\\\\\\\sf{=\sqrt{\dfrac{(1+sinA)^2}{cos^2A} } }\\\\\\\sf{=\sqrt{\dfrac{(1+sinA)^2}{(cosA)^2} } }\\\\\\\sf{=\sqrt{\bigg(\dfrac{1+sinA}{cosA} \bigg)^2 } }\\\\\\\sf{=\dfrac{1+sinA}{cosA} } \\\\\\\sf{=\dfrac{1}{cosA}+\dfrac{sinA}{cosA} }\\\\\\\sf{=secA+tanA}$

Therefore, the answer is (B) sec A + tan A.

_______________________

Identities used :--

1 - sin²A = cos²A

1/cos A = sec A

sin A/cos A = tan A