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Let 3sin^-1(log2x) + cos^-1(log2y) = π/2 and sin^-1(log2x) + 2cos^-1(log2y) = 11π/6 then

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Let 3sin-1(log2x) + cos-1(log2y) = π/2 and sin-1(log2x) + 2cos-1(log2y) = 11π/6 then find the value of (1/x2 + 1/y2 + 2)

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a = sin-1(log2x) and b = cos-1(log2x)

The given equations reduces to

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