how is 1/(cosec^2-sec^2) = sin^2/(1-sin^4)

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Arun Kumar · Answer

its not put theta =30 left part is 3/8 right part is 4/15

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how is 1/(cosec^2-sec^2) = sin^2/(1-sin^4)

how is 1/(cosec^2-sec^2) = sin^2/(1-sin^4)

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how is 1/(cosec^2-sec^2) = sin^2/(1-sin^4)

how is 1/(cosec^2-sec^2) = sin^2/(1-sin^4)

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