# ${\text{Prove that }}\tan {15^ \circ } + \tan {30^ \circ } + \tan {15^ \circ }\tan {30^ \circ } = 1$

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{\text{We know that tan4}}{{\text{5}}^ \circ } = 1 \\

{\text{We can write tan4}}{{\text{5}}^ \circ } = \tan ({30^ \circ } + {15^ \circ }) \\

{\text{We also know that tan(}}A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} \\

{\text{By using this we can write }} \\

\Rightarrow \tan ({30^ \circ } + {15^ \circ }) = \dfrac{{\tan {{30}^ \circ } + \tan {{15}^ \circ }}}{{1 - \tan {{30}^ \circ }\tan {{15}^ \circ }}} = \tan {45^ \circ } = 1 \\

{\text{By solving above equation}} \\

\tan {30^ \circ } + \tan {15^ \circ } = 1 - \tan {30^ \circ }\tan {15^ \circ } \\

{\text{By rearranging the equation we get the result}} \\

\tan {30^ \circ } + \tan {15^ \circ } + \tan {30^ \circ }\tan {15^ \circ } = 1{\text{ }}proved \\

{\text{Note: - In such type of question always try to apply the formula of tan(}}A + B){\text{ or tan(}}A - B) \\

{\text{ and put the angles that are given in question so you can prove it}}{\text{.}} \\

$

\\

{\text{We know that tan4}}{{\text{5}}^ \circ } = 1 \\

{\text{We can write tan4}}{{\text{5}}^ \circ } = \tan ({30^ \circ } + {15^ \circ }) \\

{\text{We also know that tan(}}A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} \\

{\text{By using this we can write }} \\

\Rightarrow \tan ({30^ \circ } + {15^ \circ }) = \dfrac{{\tan {{30}^ \circ } + \tan {{15}^ \circ }}}{{1 - \tan {{30}^ \circ }\tan {{15}^ \circ }}} = \tan {45^ \circ } = 1 \\

{\text{By solving above equation}} \\

\tan {30^ \circ } + \tan {15^ \circ } = 1 - \tan {30^ \circ }\tan {15^ \circ } \\

{\text{By rearranging the equation we get the result}} \\

\tan {30^ \circ } + \tan {15^ \circ } + \tan {30^ \circ }\tan {15^ \circ } = 1{\text{ }}proved \\

{\text{Note: - In such type of question always try to apply the formula of tan(}}A + B){\text{ or tan(}}A - B) \\

{\text{ and put the angles that are given in question so you can prove it}}{\text{.}} \\

$

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