Question

prove that √{(1-sinA)/(1+sinA)}=secA-tanA​

Answer 1

Answer:

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Answer 2

SOLUTION ::

LHS

= √(1 - sinA)/√(1 + sinA)

= √(1 - sinA) × √(1 - sinA)/√(1 -sinA)×√(1 +sinA)

= √(1 - SinA)²/√(1 -sin²A)

= (1 - sinA)/√cos²A

= (1 - sinA)/cosA

= 1/cosA - sinA/cosA

= SecA - tanA

= RHS

Hence the proof follows

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