If tan A =1\\3 and tanB=1/3 and tan(A+B)=tanA+tanB/1-tanA tanB show th

1.	If tan A =1\\3 and tanB=1/3 and tan(A+B)=tanA+tanB/1-tanA tanB show that A+B=45°
Answer» Given\xa0{tex} \\tan A = \\frac{1}{2},\\tan B = \\frac{1}{3}{/tex}and\xa0{tex}\\tan (A + B) = \\frac{{\\tan A + \\tan B}}{{1 - \\tan A\\tan B}}{/tex}{tex}\\Rightarrow \\tan (A + B) = \\frac{{\\frac{1}{2} + \\frac{1}{3}}}{{1 - \\frac{1}{2} \\times \\frac{1}{3}}}{/tex}{tex}\\Rightarrow \\tan (A + B) = \\frac{{\\frac{5}{6}}}{{\\frac{5}{6}}}{/tex}{tex}\\Rightarrow \\tan (A + B) = 1{/tex}{tex}\\Rightarrow \\tan (A + B) = \\tan 45^\\circ {/tex}{tex}\\Rightarrow A + B = 45^\\circ {/tex}

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