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प्रश्न
Proved that `(1 + secA)/secA = (sin^2A)/(1 - cos A)`.
उत्तर
L.H.S. = `(1 + sec A)/sec A`
= `(1 + 1/ cos A)/(1/cos A)`
= `((cos A + 1)/cos A)/(1/cos A)`
= 1 + cos A = `((1 + cos A))/1 xx ((1 - cos A))/((1 - cos A))`
= `(1 - cos^2 A)/(1 - cos A)`
`\implies (sin^2 A)/(1 - cos A)` = R.H.S. ...(∵ sin2 A + cos2 A = 1)
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