Dear Student, tan-1x+cot-1y=tan-13⇒tan-1x+tan-11y=tan-13using formula tan-1a+tan-1b=tan-1a+b1-ab⇒tan-1x+1y1-x×1y=tan-13⇒tan-1xy+1y-x=tan-13taking tan on both sidestantan-1xy+1y-x=tantan-13⇒xy+1y-x=3⇒xy+1=3y-x⇒xy+1=3y-3x⇒xy-3y=-1-3x⇒yx-3=-1+3x⇒y=3x+13-x --1for y to be positive denominator should be positivei e 3-x>0x<3hence x=1,2putting values of x =1 and 2 in equation 1for x=1, y=3+13-1⇒2for x=2, y=3×2+13-2⇒7Hence there are 2 positive integral solutionsx=1,y=2and x=2,y=7 Regards,

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Dear Student,

\tan^{- 1} x + co t^{- 1} y = \tan^{- 1} 3 \Rightarrow \tan^{- 1} x + \tan^{- 1} (\frac{1}{y}) = \tan^{- 1} 3 u \sin g f o r m u l a \tan^{- 1} a + \tan^{- 1} b = \tan^{- 1} (\frac{a + b}{1 - a b}) \Rightarrow \tan^{- 1} (\frac{x + \frac{1}{y}}{1 - x \times \frac{1}{y}}) = \tan^{- 1} 3 \Rightarrow \tan^{- 1} (\frac{x y + 1}{y - x}) = \tan^{- 1} 3 t a k i n g \tan o n b o t h s i d e s \tan [\tan^{- 1} (\frac{x y + 1}{y - x})] = \tan [\tan^{- 1} 3] \Rightarrow \frac{x y + 1}{y - x} = 3 \Rightarrow x y + 1 = 3 (y - x) \Rightarrow x y + 1 = 3 y - 3 x \Rightarrow x y - 3 y = - 1 - 3 x \Rightarrow y [x - 3] = - [1 + 3 x] \Rightarrow y = \frac{3 x + 1}{3 - x} - - (1) f o r y t o b e p o s i t i v e d e n o m i n a t o r s h o u l d b e p o s i t i v e i . e . 3 - x > 0 x < 3 h e n c e x = 1, 2 p u t t i n g v a l u e s o f x = 1 a n d 2 i n e q u a t i o n 1 f o r x = 1, y = \frac{3 + 1}{3 - 1} \Rightarrow 2 f o r x = 2, y = \frac{3 \times 2 + 1}{3 - 2} \Rightarrow 7 H e n c e t h e r e a r e 2 p o s i t i v e i n t e g r a l s o l u t i o n s x = 1, y = 2 a n d x = 2, y = 7

Regards,

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