The number of positive integral solutions of | vedclass

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(C) Given the equation: $	an ^{-1} x + \cos ^{-1}\left(\frac{y}{\sqrt{1+y^2}}ight) = \sin ^{-1}\left(\frac{3}{\sqrt{10}}ight)$.We know that $\cos

Vedclass · Accepted Answer

$2$

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(C) Given the equation: $	an ^{-1} x + \cos ^{-1}\left(\frac{y}{\sqrt{1+y^2}}ight) = \sin ^{-1}\left(\frac{3}{\sqrt{10}}ight)$.We know that $\cos ^{-1}\left(\frac{y}{\sqrt{1+y^2}}ight) = 	an ^{-1}\left(\frac{1}{y}ight)$ for $y > 0$.Also,$\sin ^{-1}\left(\frac{3}{\sqrt{10}}ight) = 	an ^{-1}(3)$.Substituting these into the equation,we get: $	an ^{-1} x + 	an ^{-1}\left(\frac{1}{y}ight) = 	an ^{-1}(3)$.Using the formula $	an ^{-1} A + 	an ^{-1} B = 	an ^{-1}\left(\frac{A+B}{1-AB}ight)$,we have: $	an ^{-1}\left(\frac{x + \frac{1}{y}}{1 - \frac{x}{y}}ight) = 	an ^{-1}(3)$.This implies $\frac{xy + 1}{y - x} = 3$,which simplifies to $xy + 1 = 3y - 3x$.Rearranging the terms: $xy + 3x - 3y = -1$.Adding $9$ to both sides to factor: $x(y+3) - 3(y+3) + 9 = -1 + 9$.$(x-3)(y+3) = 8$.Since $x$ and $y$ are positive integers,$x \ge 1$ and $y \ge 1$. Thus,$y+3 \ge 4$.The possible factors of $8$ such that $y+3 \ge 4$ are:$1) y+3 = 4 \implies y=1$. Then $x-3 = 2 \implies x=5$.$2) y+3 = 8 \implies y=5$. Then $x-3 = 1 \implies x=4$.Both $(5, 1)$ and $(4, 5)$ are positive integral solutions.Therefore,there are $2$ solutions.

The number of positive integral solutions of $\tan ^{-1} x+\cos ^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right)=\sin ^{-1}\left(\frac{3}{\sqrt{10}}\right)$ are

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