Question

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

Answer 1

Hint: We will be using the concept of trigonometric identities to solve the problem. We will using the fact that $\tan \left( 45{}^\circ \right)=1$ and these we will write \[45{}^\circ \ as\ 30{}^\circ +15{}^\circ \] and apply the trigonometric formula that $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$.

Complete step-by-step answer:

Now, we have to prove that $\tan 15{}^\circ +\tan 30{}^\circ +\tan 15{}^\circ \tan 30{}^\circ =1$.

Now, we know that,

$\tan \left( 45{}^\circ \right)=1$

Now, we will write \[45{}^\circ \ as\ 30{}^\circ +15{}^\circ \]. So, we have,

$\tan \left( 30{}^\circ +15{}^\circ \right)=1$

Now, we know that,

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

So, now using this we have that,

$\dfrac{\tan 30+\tan 15}{1-\tan 30\tan 15}=1$

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

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

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

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

Now, we have LHS = RHS.

Hence Proved.

Note: To solve these type of questions it is important to note that we have used a fact that $\tan 45{}^\circ =1$, then break \[45{}^\circ =30{}^\circ +15{}^\circ \] and applied the formula for $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ to complete the proof.