Question

prove that tan(45+theta)=1+tan theta/1-tan theta

Answer 1

SOLUTION :

TO PROVE

$\displaystyle \sf{} \tan( {45}^{ \circ} + \theta) = \frac{1 + \tan \theta}{ 1 - \tan \theta\: }$

FORMULA TO BE IMPLEMENTED

We are aware of the Trigonometric formula that

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

EVALUATION

Here we know that

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

$\sf{}Now \: taking \: A = {45}^{ \circ} \: \: and \: \: B = \theta \: \: we \: get$

$\displaystyle \sf{} \tan( {45}^{ \circ} + \theta) = \frac{ \tan {45}^{ \circ} + \tan \theta}{ 1 - \tan {45}^{ \circ} \tan \theta\: }$

$\implies \displaystyle \sf{} \tan( {45}^{ \circ} + \theta) = \frac{1 + \tan \theta}{ 1 - \tan \theta\: } \: \: (\because \: \tan {45}^{ \circ} = 1 \: )$

Hence proved

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