If $\cos^{-1}(\frac{y}{b})=\log(\frac{x}{n})^x$ , then x 2 y 2 + xy 1 =

The correct option is (B) : -n 2 y.

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If $\cos^{-1}(\frac{y}{b})=\log(\frac{x}{n})^x$, then x²y₂ + xy¹ =

Updated On: Sep 15, 2024

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The correct option is (B) : -n²y.

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