Question

(sin A tan A)/(1-cos A) = 1 + sec A
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Answer 1

LHS

(sin A tan A)/(1-cos A)

(sinA.sinA/cosA)/1-cosA [ tanA =sinA /cos A]

(sin^2A /cosA)/1-cosA

(sin^2A /cosA)× (1/1-cosA)

sin^2A/cosA(1-cosA)

(1-cos^2A)/cosA(1-cosA) [ sin^2+cos^2= 1 ]

(1+cosA)(1-cosA)/cosA(1-cosA)

(1+cosA)/cosA

(1/cosA)+cosA/cosA

secA+1

1+secA= RHS

Hence proved..