Question

Prove that: $\tan {70^ \circ } - \tan {50^ \circ } + \tan {10^ \circ } = \sqrt 3$.

Answer 1

Hint : To find the value of $\tan {70^ \circ } - \tan {50^ \circ } + \tan {10^ \circ }$, we will first write the angles in terms of sum of known angles then we will expand it using trigonometric formulas. On simplification and using trigonometric triple angle identity we will further simplify by putting the value of known angles to find the result.
Some trigonometric formulas that we will use are as follows:
$\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
$\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$

Complete step-by-step answer :
We have to find the value of $\tan {70^ \circ } - \tan {50^ \circ } + \tan {10^ \circ }$. For this we will start with writing the angles in terms of standard known angles.
We will write ${70^ \circ }$ as $\left( {{{60}^ \circ } + {{10}^ \circ }} \right)$ and ${50^ \circ }$ as $\left( {{{60}^ \circ } - {{10}^ \circ }} \right)$ because ${60^ \circ }$ is a standard angle and we know its values for different trigonometric functions.
Some trigonometric formulas that we will use are as follows:
$\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
$\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
Also, we have $\tan {60^ \circ } = \sqrt 3$ and $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$.
We can write the given expression as
\Rightarrow $$$$\tan {70^ \circ } - \tan {50^ \circ } + \tan {10^ \circ }
$= \tan \left( {{{60}^ \circ } + {{10}^ \circ }} \right) - \tan \left( {{{60}^ \circ } - {{10}^ \circ }} \right) + \tan {10^ \circ }$
Using the formula, we can write
$= \dfrac{{\tan {{60}^ \circ } + \tan {{10}^ \circ }}}{{1 - \left( {\tan {{60}^ \circ } \times \tan {{10}^ \circ }} \right)}} - \dfrac{{\tan {{60}^ \circ } - \tan {{10}^ \circ }}}{{1 + \left( {\tan {{60}^ \circ } \times \tan {{10}^ \circ }} \right)}} + \tan {10^ \circ }$
Putting the value of $\tan {60^ \circ }$, we get
$= \dfrac{{\sqrt 3 + \tan {{10}^ \circ }}}{{1 - \left( {\sqrt 3 \times \tan {{10}^ \circ }} \right)}} - \dfrac{{\sqrt 3 - \tan {{10}^ \circ }}}{{1 + \left( {\sqrt 3 \times \tan {{10}^ \circ }} \right)}} + \tan {10^ \circ }$
On rewriting we get
$= \dfrac{{\sqrt 3 + \tan {{10}^ \circ }}}{{1 - \sqrt 3 \tan {{10}^ \circ }}} - \dfrac{{\sqrt 3 - \tan {{10}^ \circ }}}{{1 + \sqrt 3 \tan {{10}^ \circ }}} + \tan {10^ \circ }$
On taking LCM, we get
$= \dfrac{{\left( {\sqrt 3 + \tan {{10}^ \circ }} \right)\left( {1 + \sqrt 3 \tan {{10}^ \circ }} \right) - \left( {\sqrt 3 - \tan {{10}^ \circ }} \right)\left( {1 - \sqrt 3 \tan {{10}^ \circ }} \right) + \tan {{10}^ \circ }\left( {1 - \sqrt 3 \tan {{10}^ \circ }} \right)\left( {1 + \sqrt 3 \tan {{10}^ \circ }} \right)}}{{\left( {1 - \sqrt 3 \tan {{10}^ \circ }} \right)\left( {1 + \sqrt 3 \tan {{10}^ \circ }} \right)}}$
On multiplication and using the formula $(a - b)(a + b) = {a^2} - {b^2}$, we get
$= \dfrac{{\left( {\sqrt 3 + 3\tan {{10}^ \circ } + \tan {{10}^ \circ } + \sqrt 3 {{\tan }^2}{{10}^ \circ }} \right) - \left( {\sqrt 3 - 3\tan {{10}^ \circ } - \tan {{10}^ \circ } + \sqrt 3 {{\tan }^2}{{10}^ \circ }} \right) + \left( {\tan {{10}^ \circ } - 3{{\tan }^3}{{10}^ \circ }} \right)}}{{\left( {1 - 3{{\tan }^2}{{10}^ \circ }} \right)}}$
On simplification, we get
$= \dfrac{{\sqrt 3 + 3\tan {{10}^ \circ } + \tan {{10}^ \circ } + \sqrt 3 {{\tan }^2}{{10}^ \circ } - \sqrt 3 + 3\tan {{10}^ \circ } + \tan {{10}^ \circ } - \sqrt 3 {{\tan }^2}{{10}^ \circ } + \tan {{10}^ \circ } - 3{{\tan }^3}{{10}^ \circ }}}{{1 - 3{{\tan }^2}{{10}^ \circ }}}$
On simplification, we get
$= \dfrac{{9\tan {{10}^ \circ } - 3{{\tan }^3}{{10}^ \circ }}}{{1 - 3{{\tan }^2}{{10}^ \circ }}}$
On taking $3$ common from the numerator we get
$= 3 \times \dfrac{{\left( {3\tan {{10}^ \circ } - {{\tan }^3}{{10}^ \circ }} \right)}}{{1 - 3{{\tan }^2}{{10}^ \circ }}} - - - (1)$
As we know, $\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}$.
Using this we can write $(1)$
$= 3 \times \tan \left( {3 \times {{10}^ \circ }} \right)$
$= 3\tan {30^ \circ }$
Using $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$, we get
$\Rightarrow \tan {70^ \circ } - \tan {50^ \circ } + \tan {10^ \circ } = 3 \times \dfrac{1}{{\sqrt 3 }}$
On simplification,
$\Rightarrow \tan {70^ \circ } - \tan {50^ \circ } + \tan {10^ \circ } = \sqrt 3$
Hence, proved that $\tan {70^ \circ } - \tan {50^ \circ } + \tan {10^ \circ } = \sqrt 3$.

Note : Here, we have written $\tan {70^ \circ }$ as $\tan \left( {{{60}^ \circ } + {{10}^ \circ }} \right)$ and $\tan {50^ \circ }$ as $\tan \left( {{{60}^ \circ } - {{10}^ \circ }} \right)$ because we have known values of only few trigonometric angles. So, we have written it in a way that it can be splitted into known angles and we can substitute direct values of these angles which will simplify the equation. Also note that, we have not changed $\tan {10^ \circ }$ because there is no way to write the sum of this angle in a way that it can give standard angles where we can put direct results.