Question

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

Answer 1

$$\begin{gathered} \text{We know that tan4}{\text{5}}^{\circ }=1 \\ \text{We can write tan4}{\text{5}}^{\circ }=\tan ({30}^{\circ }+{15}^{\circ }) \\ \text{We also know that tan(}A+B)={\displaystyle \frac{\tan A+\tan B}{1-\tan A\tan B}} \\ \text{By using this we can write } \\ \Rightarrow \tan ({30}^{\circ }+{15}^{\circ })={\displaystyle \frac{\tan {30}^{\circ }+\tan {15}^{\circ }}{1-\tan {30}^{\circ }\tan {15}^{\circ }}}=\tan {45}^{\circ }=1 \\ \text{By solving above equation} \\ \tan {30}^{\circ }+\tan {15}^{\circ }=1-\tan {30}^{\circ }\tan {15}^{\circ } \\ \text{By rearranging the equation we get the result} \\ \tan {30}^{\circ }+\tan {15}^{\circ }+\tan {30}^{\circ }\tan {15}^{\circ }=1proved \\ \text{Note: - In such type of question always try to apply the formula of tan(}A+B)\text{ or tan(}A-B) \\ \text{ and put the angles that are given in question so you can prove it}\text{.} \end{gathered}$$