VerifiedHint: Here use trigonometric identities and formulas on the LHS part of the equation and simplify using basic trigonometric angles to prove LHS=RHS.
Complete step-by-step answer:
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.
