If tan=1/3 and tan b=1/2 . …
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Neeraj Sharma 7 years, 1 month ago
{tex}Given\,that{/tex}
{tex}\tan A = {1 \over 3}\,and\,\tan B = {1 \over 2}{/tex}
{tex}Now\,\,\tan \left( {A + B} \right) = {{\tan A + \tan B} \over {1 - tan A \cdot \tan B}}{/tex}
{tex} = {{\left( {{1 \over 3} + {1 \over 2}} \right)} \over {\left( {1 - {1 \over 3} \cdot {1 \over 2}} \right)}}{/tex}
{tex} = {{\left( {{5 \over 6}} \right)} \over {\left( {{5 \over 6}} \right)}}{/tex}
{tex}\tan \left( {A + B} \right) = 1{/tex}
{tex}A + B = {45^ \circ }{/tex}
{tex}Now\,Sin2\left( {A + B} \right) = \sin \left( {2 \times {{45}^ \circ }} \right){/tex}
{tex} = \sin {90^ \circ }{/tex}
{tex}\sin 2\left( {A + B} \right) = 1{/tex}
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