If x and y are positive integers satisfying tan-1 1x -tan-1 1y -tan-1 17 - then the number of ordered pairs of x- y is

Tan^ 1 (1x )+tan^ 1 ( 1y )=tan^ 1 ( 17) ⇒tan^ 1 (1x+1y1 1xy )=tan^ 1 ( 17 ) ⇒x+yxy 1=17 ⇒ 7x+7y=xy 1 ⇒ x(y 7)=7y+1 ⇒ x=7y+1y 7=7(y 7+7)+1y 7 ⇒ x=7+50y 7 Since x ...

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Standard XII

Mathematics

Range of Trigonometric Expressions

If x and y ar...

Question

If $x$ and $y$ are positive integers satisfying ${tan}^{- 1} (\frac{1}{x}) + {tan}^{- 1} (\frac{1}{y}) = {tan}^{- 1} (\frac{1}{7}),$ then the number of ordered pairs of $(x, y)$ is

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Solution

${tan}^{- 1} (\frac{1}{x}) + {tan}^{- 1} (\frac{1}{y}) = {tan}^{- 1} (\frac{1}{7})$
$\Rightarrow {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{1}{x} + \frac{1}{y}}{1 - \frac{1}{x y}} ⎞ ⎟ ⎟ ⎟ ⎠ = {tan}^{- 1} (\frac{1}{7})$
$\Rightarrow \frac{x + y}{x y - 1} = \frac{1}{7}$
$\Rightarrow 7 x + 7 y = x y - 1$
$\Rightarrow x (y - 7) = 7 y + 1$
$\Rightarrow x = \frac{7 y + 1}{y - 7} = \frac{7 (y - 7 + 7) + 1}{y - 7}$
$\Rightarrow x = 7 + \frac{50}{y - 7}$
Since $x$ is a positive integer, $y$ can take the values $8, 9, 12, 17, 32, 57.$
Hence, number of ordered pairs of $(x, y)$ is $6.$

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If x and y are positive integers satisfying tan−1(1x)+tan−1(1y)=tan−1(17), then the number of ordered pairs of (x,y) is

If $x$ and $y$ are positive integers satisfying ${tan}^{- 1} (\frac{1}{x}) + {tan}^{- 1} (\frac{1}{y}) = {tan}^{- 1} (\frac{1}{7}),$ then the number of ordered pairs of $(x, y)$ is