Question

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

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Hint: Here use trigonometric identities and formulas on the LHS part of the equation and simplify using basic trigonometric angles to prove LHS=RHS.

We know that $tan {45^ \circ } = 1$

We can write $tan {45^ \circ } = \tan ({30^ \circ } + {15^ \circ })$

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

By using this we can write

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

By solving above equation

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

By rearranging the equation we get the result

$\Rightarrow \tan {30^ \circ } + \tan {15^ \circ } + \tan {30^ \circ }\tan {15^ \circ } = 1$

Hence Proved

Note: - In such a type of question always try to apply the formula of $tan(A + B)$ or $tan(A - B)$ and put the angles that are given in question so we can prove it.