If x, y, and a are positive numbers such that xy = a^2 + 1, then show that

tan^-1 (1 / (a + x)) + tan^-1 (1 / (a + y)) = tan^-1 (1 / a)
for all acute angles. Hence deduce that
tan^-1 (1 / 2) + tan^-1 (1 / 3) = pi / 4
and
tan^-1 (1 / 3) + tan^-1 (1 / 5) + tan^-1 (1 / 7) + tan^-1 (1 / 8) = pi / 4

Let : xy = a² + 1.......................................… (1)

Put : 1/(a+x) = α and 1/(a+y) = ß.

Then :

( α + ß ) = (2a+x+y) / (a+x)(a+y) ........................................… (2)

1 - αß = [ (a+x)(a+y) - 1 ] / (a+x)(a+y)

. . . . .= [ ( a² + xy + a(x+y) - 1 ] / ()()

. . . . .= [ a² + ( a² + 1 ) + a(x+y) - 1 ] / ()() ... from (1)

. . . . .= [ 2a² + a(x+y) ] / ()()

. . . . .= a( 2a + x + y ) / ()() ........................................… (3)
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From (2) and (3),

( α + ß ) / ( 1 - αß ) = 1/a

so that

tanֿ¹ [ ( α + ß ) / ( 1 - αß ) ] = tanֿ¹ ( 1/a )

so that

tanֿ¹ α + tanֿ¹ ß = tanֿ¹ ( 1/a ),

that is,

tanֿ¹ [ 1/ (a+x) ] + tanֿ¹ [ a+y) ] = tanֿ¹ ( 1/a ) ............................ (4)

Hence, the result.
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Deduction-1 :

Put a = 1, x = 1 and y = 2 in (4).

Also : tanֿ¹ (1) = π/4.
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Deduction-2 :

Use the method shown in

Deduction-1 above.
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