Question

Prove that the value of $\tan {{70}^{\circ }}-\tan {{20}^{\circ }}=2\tan {{50}^{\circ }}$.

Answer 1

Hint: Here, we can write ${{70}^{\circ }}={{20}^{\circ }}+{{50}^{\circ }}$, then apply tan on both the sides. After that apply the formulas.

Complete step-by-step answer:
$\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$, $\tan A=\cot ({{90}^{\circ }}-A)$ and $\tan A\cot A=1$.

Here, we have to prove that $\tan {{70}^{\circ }}-\tan {{20}^{\circ }}=2\tan {{50}^{\circ }}$.
For that first, we have to write:
${{70}^{\circ }}={{20}^{\circ }}+{{50}^{\circ }}$
Now, by applying tan on both the sides we get,
$\tan {{70}^{\circ }}=\tan ({{20}^{\circ }}+\tan {{50}^{\circ }})\text{ }.....\text{ (1)}$
RHS is in the form of $\tan (A+B)$. We have a formula for $\tan (A+B)$, the formula is given by:
$\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ .
In our equation (1) we have $A={{20}^{\circ }}$, $B={{50}^{\circ }}$ and $A+B={{70}^{\circ }}$. Now by applying the above formula to equation (1) we obtain:
$\tan {{70}^{\circ }}=\dfrac{\tan {{20}^{\circ }}+\tan {{50}^{\circ }}}{1-\tan {{20}^{\circ }}\tan {{50}^{\circ }}}$
By cross multiplication our equation becomes,
$\tan {{70}^{\circ }}\left( 1-\tan {{20}^{\circ }}\tan {{50}^{\circ }} \right)=\tan {{20}^{\circ }}+\tan {{50}^{\circ }}$
In the next step, we have to multiply $\tan {{70}^{\circ }}$with $\left( 1-\tan {{20}^{\circ }}\tan {{50}^{\circ }} \right)$, we get the equation:
$\tan {{70}^{\circ }}-\tan {{70}^{\circ }}\tan {{20}^{\circ }}\tan {{50}^{\circ }}=\tan {{20}^{\circ }}+\tan {{50}^{\circ }}\text{ }.....\text{ (2)}$
Next, we have to check that if there is any cancellation possible.
We are familiar with the formula that,
$\tan A=\cot ({{90}^{\circ }}-A)$
We can apply the above formula in equation (2) for further cancellation.
The formula can be applied for either $\tan {{20}^{\circ }}$ or $\tan {{70}^{\circ }}$.Here we are applying for $\tan {{20}^{\circ }}$. i.e. by applying the above formula we get,
$\begin{aligned} & \tan {{20}^{\circ }}=\cot \left( {{90}^{\circ }}-{{20}^{\circ }} \right) \\\ & \tan {{20}^{\circ }}=\cot {{70}^{\circ }}\text{ }....\text{ (3)} \\\ \end{aligned}$
By substituting equation (3) in equation (2) we get:
$\tan {{70}^{\circ }}-\tan {{70}^{\circ }}\cot {{70}^{\circ }}\tan {{50}^{\circ }}=\tan {{20}^{\circ }}+\tan {{50}^{\circ }}\text{ }....\text{ (4)}$
We also know that $\tan A\cot A=1$.
In equation (4) we have $A={{70}^{\circ }}$, so we can say that $\tan {{70}^{\circ }}\cot {{70}^{\circ }}=1$
Therefore, our equation (4) becomes:
$\begin{aligned} & \tan {{70}^{\circ }}-1\times \tan {{50}^{\circ }}=\tan {{20}^{\circ }}+\tan {{50}^{\circ }}\text{ } \\\ & \tan {{70}^{\circ }}-\tan {{50}^{\circ }}=\tan {{20}^{\circ }}+\tan {{50}^{\circ }}\text{ } \\\ \end{aligned}$
Next, by taking $\tan {{20}^{\circ }}$to the left side, $\tan {{20}^{\circ }}$ becomes $-\tan {{20}^{\circ }}$. Hence, our equation becomes:
$\tan {{70}^{\circ }}-\tan {{50}^{\circ }}-\tan {{20}^{\circ }}=\tan {{50}^{\circ }}\text{ }$
In the next step take $-\tan {{50}^{\circ }}$ to the right side, then $-\tan {{50}^{\circ }}$ becomes $\tan {{50}^{\circ }}$. So, we obtain the equation:
$\begin{aligned} & \tan {{70}^{\circ }}-\tan {{20}^{\circ }}=\tan {{50}^{\circ }}+\tan {{50}^{\circ }}\text{ } \\\ & \tan {{70}^{\circ }}-\tan {{20}^{\circ }}=2\tan {{50}^{\circ }} \\\ \end{aligned}$
Hence, we have proved that $\tan {{70}^{\circ }}-\tan {{20}^{\circ }}=2\tan {{50}^{\circ }}$.

Note: To solve this problem we should be familiar with the trigonometric formulas. Here, alternatively you can directly apply the formula for $\tan A-\tan B.$