Question

Prove that: $\tan {{70}^{\circ }}=\tan {{20}^{\circ }}+2\tan {{50}^{\circ }}$.

Answer 1

Hint: We have a tangent of angle as: $\tan {{70}^{\circ }}$ . We can write this expression as: $\tan {{\left( 20+50 \right)}^{\circ }}$
Now, by addition rule of tangent of angles, i.e. $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A.\tan B}$ solve the expression $\tan {{\left( 20+50 \right)}^{\circ }}$.Then check if you can convert any angles into cotangent and cancel out the terms to get the final expression $\tan {{70}^{\circ }}=\tan {{20}^{\circ }}+2\tan {{50}^{\circ }}$.

Complete step by step answer:
We have the following expression: $\tan {{70}^{\circ }}......(1)$
We can write equation (1) as:
$\tan {{70}^{\circ }}=\tan {{\left( 20+50 \right)}^{\circ }}......(2)$
Now, by using addition rule of tangent of angles, i.e. $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A.\tan B}$ for equation (2), we get:
$\tan {{70}^{\circ }}=\dfrac{\tan {{20}^{\circ }}+\tan {{50}^{\circ }}}{1-\tan {{20}^{\circ }}.\tan {{50}^{\circ }}}......(3)$
We can also write equation (3) as:
\[\tan {{70}^{\circ }}\left( 1-\tan {{20}^{\circ }}.\tan {{50}^{\circ }} \right)=\tan {{20}^{\circ }}+\tan {{50}^{\circ }}......(4)\]
Now, expand equation (4), we get:
\[\left( \tan {{70}^{\circ }}-\tan {{20}^{\circ }}.\tan {{50}^{\circ }}\tan {{70}^{\circ }} \right)=\tan {{20}^{\circ }}+\tan {{50}^{\circ }}......(5)\]
We can write equation (5) as:
\[\tan {{70}^{\circ }}=\tan {{20}^{\circ }}+\tan {{50}^{\circ }}+\tan {{20}^{\circ }}.\tan {{50}^{\circ }}\tan {{70}^{\circ }}......(6)\]
We can write $\tan {{20}^{\circ }}=\tan {{\left( 90-70 \right)}^{\circ }}$
We get:
\[\tan {{70}^{\circ }}=\tan {{20}^{\circ }}+\tan {{50}^{\circ }}+\tan {{\left( 90-70 \right)}^{\circ }}.\tan {{50}^{\circ }}\tan {{70}^{\circ }}......(7)\]
We know that $\tan \left( 90-\theta \right)=\cot \theta $
So, we get:
\[\tan {{70}^{\circ }}=\tan {{20}^{\circ }}+\tan {{50}^{\circ }}+\cot {{70}^{\circ }}.\tan {{50}^{\circ }}\tan {{70}^{\circ }}......(8)\]
Also, we know that: $\tan \theta .\cot \theta =1$
So, we get:
\[\begin{align}
  & \Rightarrow \tan {{70}^{\circ }}=\tan {{20}^{\circ }}+\tan {{50}^{\circ }}+\tan {{50}^{\circ }} \\
 & \Rightarrow \tan {{70}^{\circ }}=\tan {{20}^{\circ }}+2\tan {{50}^{\circ }} \\
\end{align}\]
Hence proved that $\tan {{70}^{\circ }}=\tan {{20}^{\circ }}+2\tan {{50}^{\circ }}$.

Note:
Always remember that, whenever we are given to solve an expression having tangent of angles, try to solve it by addition rule for tangent of angles, i.e. $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A.\tan B}$ or subtraction rule of tangent of angles, i.e. $\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A.\tan B}$ . Also, we can turn angles into cotangent and cancel out the terms to make the expression simpler.