Question

Underoot1-sinA/1+sinA =secA - tanA​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

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

Squaring both sides:

$\frac{1-sinA}{1+sinA} =(secA-tanA)^2$

Solving LHS:

$(secA-tanA)^2$

$sec^2A+tan^2A-2tanAsecA$

$\frac{1}{cos^2A} +\frac{sin^2A}{cos^2A} -2(\frac{1}{cosA} )(\frac{sinA}{cosA} )$

$\frac{1+sin^2A-2sinA}{cos^A}$

$\frac{(1-sinA)^2}{1-sin^2A}$

$\frac{(1-sinA)(1-sinA)}{(1-sinA)(1+sinA)}$

$\frac{1-sinA}{1+sinA}$

Since, LHS=RHS

Hence proved