In a ` A B C ,`, if C is a right angle, then `tan^(-1)(a/(b+c))+tan^(-1)(b/(c+a))=``pi/3`(b) `pi/4`(c) `(5pi)/2`(d) `pi/6`

In a ` A B C ,`, if C is a right angle, then `tan^(-1)(a/(b+c))+tan^(-

1.	In a ` A B C ,`, if C is a right angle, then `tan^(-1)(a/(b+c))+tan^(-1)(b/(c+a))=``pi/3`(b) `pi/4`(c) `(5pi)/2`(d) `pi/6`
Answer» `tan^-1 x + tan^-1 y = tan^-1 ((x+y)/(1-xy))` so, `tan^-1 (a/(b+c)) + tan^-1(b/(a+c))` `tan^-1((a/(b+c) + b/(a+c))/(1- (ab)/((a+c)(b+c))))` `= tan^-1[ (a^2 + ac + b^2 + bc)/(ab + bc + ac + c^2 - ab)]` `= tan^-1[(a^2+ b^2 + ac+bc)/(bc + ac+c^2)]` so, `/_ c= pi/2` `cos c = (a^2 + b^2 - c^2)/(2ab)=0` `a^2 + b^2 = c^2` `tan^-1 [ (bc+ac+c^2)/(bc + ac+c^2)]` `= tan^-1 1 = pi/4` Answer

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In a ` A B C ,`, if C is a right angle, then `tan^(-1)(a/(b+c))+tan^(-1)(b/(c+a))=``pi/3`(b) `pi/4`(c) `(5pi)/2`(d) `pi/6`
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