If tan A =1\3 and tanB=1/3 …
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Posted by Vikash Yadav 4 years, 10 months ago
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Sia ? 4 years, 10 months ago
Given {tex} \tan A = \frac{1}{2},\tan B = \frac{1}{3}{/tex}
and {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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