Question

value of(1+tan13)(1+tan32)+(1+tan12)(1+tan33)
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Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

$\sf{If \: \: A + B = {45}^{ \circ} } \: \: then$

$\sf{\tan( A + B )= \tan {45}^{ \circ} } \:$

$\displaystyle \: \implies \: \sf{ \frac{ \tan A + \tan B \: }{1 - \tan A \tan B\: } = 1 } \:$

$\displaystyle \: \implies \: \sf{ \tan A + \tan B \: = 1 - \tan A \tan B\: } \:$

$\displaystyle \: \implies \: \sf{ \tan A + \tan B \: + \tan A \tan B = 1\: } \:$

$\displaystyle \: \implies \: \sf{1 + \tan A + \tan B \: + \tan A \tan B = 2\: } \:$

$\displaystyle \: \implies \: \sf{(1 + \tan A) + \tan B( \: 1+ \tan A )= 2\: } \:$

$\displaystyle \: \implies \: \sf{(1 + \tan A) (1+ \tan B )= 2\: } \:$

TO DETERMINE

$\displaystyle \: \: \sf{(1 + \tan {13}^{ \circ} ) (1+ \tan {32}^{ \circ} ) + \:(1 + \tan {12}^{ \circ} ) (1+ \tan {33}^{ \circ} ) } \:$

CALCULATION

$\sf{ \:Since \: \: {13}^{ \circ} + {32}^{ \circ} = {45}^{ \circ}\: }$

So by the above mentioned formula

$\displaystyle \: \: \sf{(1 + \tan {13}^{ \circ} ) (1+ \tan {32}^{ \circ} ) = 2 } \:$

$\sf{ \:Since \: \: {12}^{ \circ} + {33}^{ \circ} = {45}^{ \circ}\: }$

So by the above mentioned formula

$\displaystyle \: \: \sf{ \:(1 + \tan {12}^{ \circ} ) (1+ \tan {33}^{ \circ} ) = 2} \:$

On addition

$\displaystyle \: \: \sf{(1 + \tan {13}^{ \circ} ) (1+ \tan {32}^{ \circ} ) + \:(1 + \tan {12}^{ \circ} ) (1+ \tan {33}^{ \circ} ) } \:$

$= 2 + 2$

$= 4$

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