Prove that: $\tan 36{}^\circ +\tan 9{}^\circ +\tan 36{}^\circ \tan 9{}^\circ =1$.

Hint: We will be using the concept of trigonometry to solve the problem. We will using the fact that $\tan 45{}^\circ =1$ then we will break \[45{}^\circ \ as\ 36{}^\circ +9{}^\circ \] and apply the formula that $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ to further simplify the solution.

Complete step-by-step answer:

Now, we have to prove that $\tan 36{}^\circ +\tan 9{}^\circ +\tan 36{}^\circ \tan 9{}^\circ =1$.

Now, we know that,

$\tan 45{}^\circ =1$

Now, in the equation we have to prove we can see that the angles are \[36{}^\circ \ and\ 9{}^\circ \]. Also we can see that, \[36{}^\circ +9{}^\circ =45{}^\circ \].

So, we have

\[\tan \left( 36{}^\circ +9{}^\circ \right)=1\]

Now, we know that,

$\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$

So, we have that,

\[\begin{align}

  & \tan \left( 36{}^\circ +9{}^\circ \right)=1 \\

 & \dfrac{\tan 36{}^\circ +\tan 9{}^\circ }{1-\tan 36{}^\circ \tan 9{}^\circ }=1 \\

\end{align}\]

Now, we will cross-multiply the terms. So, we have,

$\tan 36{}^\circ +\tan 9{}^\circ =1-\tan 36{}^\circ \tan 9{}^\circ $

Now, we will rearrange the terms. So, we have,

$\tan 36{}^\circ +\tan 9{}^\circ +\tan 36{}^\circ \tan 9{}^\circ =1$

Now, we have LHS = RHS.

Hence Proved.

Note: To solve these types of questions it is important to note that we have used a fact that $\tan 45{}^\circ =1$. Also we have splitted \[45{}^\circ \ as\ 36{}^\circ +9{}^\circ \] in the equation to be proved then we applied a formula that $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$.