If {tan ^{ - 1}}x + {cos ^{ - 1}}frac{y}{{sqr | vedclass

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

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$(1, 2)$ and $(2, 7)$

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(A) Given equation: ${	an ^{ - 1}}x + {\cos ^{ - 1}}\frac{y}{{\sqrt {1 + {y^2}} }} = {\sin ^{ - 1}}\frac{3}{{\sqrt {10} }}$We know that ${\cos ^{ - 1}}\frac{y}{{\sqrt {1 + {y^2}} }} = {	an ^{ - 1}}\frac{1}{y}$ and ${\sin ^{ - 1}}\frac{3}{{\sqrt {10} }} = {	an ^{ - 1}}3$.Substituting these,we get: ${	an ^{ - 1}}x + {	an ^{ - 1}}\frac{1}{y} = {	an ^{ - 1}}3$.Using the formula ${	an ^{ - 1}}A + {	an ^{ - 1}}B = {	an ^{ - 1}}\frac{A+B}{1-AB}$,we have: $\frac{x + \frac{1}{y}}{1 - \frac{x}{y}} = 3$.Simplifying: $\frac{xy + 1}{y - x} = 3 \implies xy + 1 = 3y - 3x \implies xy + 3x - 3y + 1 = 0$.Adding $8$ to both sides to factorize: $x(y + 3) - 3(y + 3) + 10 = 0 \implies (x - 3)(y + 3) = -10 \implies (3 - x)(y + 3) = 10$.Since $x, y$ are positive integers,we test factors of $10$: $(3-x)$ can be $1$ or $2$.If $3 - x = 1 \implies x = 2$,then $y + 3 = 10 \implies y = 7$.If $3 - x = 2 \implies x = 1$,then $y + 3 = 5 \implies y = 2$.Thus,the pairs are $(1, 2)$ and $(2, 7)$.

If ${\tan ^{ - 1}}x + {\cos ^{ - 1}}\frac{y}{{\sqrt {1 + {y^2}} }} = {\sin ^{ - 1}}\frac{3}{{\sqrt {10} }}$ and both $x$ and $y$ are positive integers,then the possible values of $(x, y)$ are:

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