OVCU 2 Call 8 4 IV. Show that tan-11 + tan-+ 2 + tan* 3 = 2(tan-11 + tan-1+tan-1)

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( tan ^{-1}(a)+tan ^{-1}(b)=tan ^{-1}left(frac{a+b}{1-a b}right) ) ( begin{aligned} therefore tan ^{-1}(1)+tan ^{-1}(2)=& tan ^{-1}left(frac{1+2}{1-(1)(2)}right)=tan ^{-1}(-3) =& 11-tan ^{-1}(3) end{aligned} ) ( pi-tan ^{-1}(3)+tan ^{-1}(3)=frac{11}{11} ) ( operatorname{ard} tan ^{-1}left(frac{1}{2}right)+tan ^{-1}left(frac{1}{3}right)=tan ^{-1}left(frac{1 / 2+1 / 3}{1-frac{1}{2}left(frac{1}{3}right)}right) ) ( =tan ^{-1}left(frac{5 / 6}{5 / 6}right)=tan ^{-1}(1) ) (0.2)( : 2left(tan ^{-1}(1)+tan ^{-1}(1)right)=2(184+pi u)=pi ) ( 12+ ) Hurce proved.

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