Question

Prove that $\tan 70^\circ = 2\tan 50^\circ + \tan 20^\circ $

Answer 1

Hint: As we know that for $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
So we can write that $\tan 70^\circ = \tan (50^\circ + 20^\circ )$
Now you can easily expand using the given formula and can prove the above equation.

Complete step-by-step answer:
As we need to prove that
$\tan 70^\circ = 2\tan 50^\circ + \tan 20^\circ $
Now we have $\tan 70^\circ $ in LHS. So as we know the formula of $\tan (A + B)$ which is
$\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
So first thing which we are seeing that in the LHS we have $\tan 70^\circ $ and we need to split it in $\tan 20^\circ {\text{ and }}\tan 50^\circ $ as
$\tan 70^\circ = \tan (50^\circ + 20^\circ )$
Now we can use $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
So we get $\tan 70^\circ = \dfrac{{\tan 20^\circ + \tan 50^\circ }}{{1 - \tan 20^\circ \tan 50^\circ }}$
Upon further simplification, we get that
$\Rightarrow$$\tan 70^\circ - \tan 70^\circ \tan 20^\circ \tan 50^\circ = \tan 20^\circ + \tan 50^\circ $
Now we can write it as
$\Rightarrow$$\tan 70^\circ = \tan 20^\circ + \tan 50^\circ + \tan 70^\circ \tan 20^\circ \tan 50^\circ $$ - - - - - (1)$
Also we know that $\tan (90 - \theta ) = \cot \theta $
$\tan \theta = \dfrac{1}{{\cot \theta }}$
So we can cross multiply
$\tan \theta \cot \theta = 1$
So we can write that
$
\Rightarrow \tan 70^\circ = \tan (90^\circ - 20^\circ ) \\
  {\text{ }} = \cot 20^\circ
 $
Now we put this value in equation (1)
$\Rightarrow$$\tan 70^\circ = \tan 20^\circ + \tan 50^\circ + \cot 20^\circ \tan 20^\circ \tan 50^\circ $
So as $\tan \theta \cot \theta = 1$
We can write that
$\Rightarrow$$\tan 20^\circ \cot 20^\circ = 1$
We get that
$\Rightarrow$$\tan 70^\circ = \tan 20^\circ + \tan 50^\circ + \tan 50^\circ $
$\Rightarrow$$\tan 70^\circ = \tan 20^\circ + 2\tan 50^\circ $
Hence proved.

Note: We should know the following conversions like
$
  \sin (90 \pm \theta ) = \cos \theta ;\sin \theta {\text{cosec}}\theta = 1 \\
  {\text{cosec}}(90 \pm \theta ) = \sec \theta ;\cos \theta \sec \theta = 1 \\
  \tan (90 \pm \theta ) = \cot \theta ;\tan \theta \cot \theta = 1 \\
 $