Question

Prove the following:
 \[\tan {10^ \circ } + \tan {70^ \circ } - \tan {50^ \circ } = \sqrt 3 \].

Answer 1

Hint: We will solve this question using the sum and difference formula of tangent function. So, we will first write \[\tan {70^ \circ }\] as \[\tan \left( {{{60}^ \circ } + {{10}^ \circ }} \right)\] and similarly we will write \[\tan {50^ \circ }\] as \[\tan \left( {{{60}^ \circ } - {{10}^ \circ }} \right)\]. Then we will use the formula for \[\tan \left( {A + B} \right)\] and \[\tan \left( {A - B} \right)\] for expanding \[\tan \left( {{{60}^ \circ } + {{10}^ \circ }} \right)\] and \[\tan \left( {{{60}^ \circ } - {{10}^ \circ }} \right)\].
Formula used:
\[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}\]
\[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A.\tan B}}\]
\[\tan 3A = \dfrac{{3\tan A - {{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}\]

Complete step by step answer:
Writing the L.H.S:
\[\tan {10^ \circ } + \tan {70^ \circ } - \tan {50^ \circ }\]
Now we will break \[\tan {70^ \circ }\] and \[\tan {50^ \circ }\]. So,
\[\tan {10^ \circ } + \tan {70^ \circ } - \tan {50^ \circ } = \tan {10^ \circ } + \tan \left( {{{60}^ \circ } + {{10}^ \circ }} \right) - \tan \left( {{{60}^ \circ } - {{10}^ \circ }} \right)\]
Now we will use the formula of \[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}\] and \[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A.\tan B}}\]
for expanding \[\tan \left( {{{60}^ \circ } + {{10}^ \circ }} \right)\] and \[\tan \left( {{{60}^ \circ } - {{10}^ \circ }} \right)\].
\[ \Rightarrow \tan {10^ \circ } + \tan \left( {{{60}^ \circ } + {{10}^ \circ }} \right) - \tan \left( {{{60}^ \circ } - {{10}^ \circ }} \right) = \tan {10^ \circ } + \left( {\dfrac{{\tan {{60}^ \circ } + \tan {{10}^ \circ }}}{{1 - \tan {{60}^ \circ }.\tan {{10}^ \circ }}}} \right) - \left( {\dfrac{{\tan {{60}^ \circ } - \tan {{10}^ \circ }}}{{1 + \tan {{60}^ \circ }.\tan {{10}^ \circ }}}} \right)\]
Now we will put the value of \[\tan {60^ \circ } = \sqrt 3 \] in the above equation. So, we will get
\[ = \tan {10^ \circ } + \left( {\dfrac{{\sqrt 3 + \tan {{10}^ \circ }}}{{1 - \sqrt 3 \tan {{10}^ \circ }}}} \right) - \left( {\dfrac{{\sqrt 3 - \tan {{10}^ \circ }}}{{1 + \sqrt 3 \tan {{10}^ \circ }}}} \right)\]
Now we will solve it further by taking the LCM of the denominator.
LCM of \[\left( {1 - \sqrt 3 \tan {{10}^ \circ }} \right)\] and \[\left( {1 + \sqrt 3 \tan {{10}^ \circ }} \right)\] is equal to \[\left( {1 - 3{{\tan }^2}{{10}^ \circ }} \right)\].
Here we have found the LCM by simply multiplying the terms using the formula:
\[\left( {a + b} \right) \times \left( {a - b} \right) = {a^2} - {b^2}\].
Now using the LCM and solving further,
\[ = \dfrac{{\tan {{10}^ \circ }\left( {1 - 3{{\tan }^2}{{10}^ \circ }} \right) + \left( {\sqrt 3 + \tan {{10}^ \circ }} \right)\left( {1 + \sqrt 3 \tan {{10}^ \circ }} \right) - \left( {\sqrt 3 - \tan {{10}^ \circ }} \right)\left( {1 - \sqrt 3 \tan {{10}^ \circ }} \right)}}{{1 - 3{{\tan }^2}{{10}^ \circ }}}\]
Solving the numerator using the distributive property:
\[ = \dfrac{{\tan {{10}^ \circ } - 3{{\tan }^3}{{10}^ \circ } + \sqrt 3 + 3\tan {{10}^ \circ } + \tan {{10}^ \circ } + \sqrt 3 {{\tan }^2}{{10}^ \circ } - \sqrt 3 + 3\tan {{10}^ \circ } + \tan {{10}^ \circ } - \sqrt 3 {{\tan }^2}{{10}^ \circ }}}{{1 - 3{{\tan }^2}{{10}^ \circ }}}\]
Now we will simplify the numerator of the above fraction by cancelling and combining the terms.
We will get,
\[ = \dfrac{{9\tan {{10}^ \circ } - 3{{\tan }^3}{{10}^ \circ }}}{{1 - 3{{\tan }^2}{{10}^ \circ }}}\]
Now we take \[3\] common in the numerator, the it becomes:
\[ = \dfrac{{3\left( {3\tan {{10}^ \circ } - {{\tan }^3}{{10}^ \circ }} \right)}}{{1 - 3{{\tan }^2}{{10}^ \circ }}}\]
Now we can observe that this is equal to \[3\tan 3A\] in which \[A = {10^ \circ }\]. So, we get;
\[ = 3\tan \left( {3 \times {{10}^ \circ }} \right)\]
\[ = 3\tan \left( {{{30}^ \circ }} \right)\]
Now we will put the value of \[\tan \left( {{{30}^ \circ }} \right) = \dfrac{1}{{\sqrt 3 }}\].
\[ = 3 \times \dfrac{1}{{\sqrt 3 }}\]
\[ = \sqrt 3 \]
Which is equal to the RHS.

Note:
One important point that one should keep in mind here is while using the formula for \[\tan \left( {A + B} \right)\] we write \[1 - \tan A.\tan B\] in the denominator and while writing the formula for \[\tan \left( {A - B} \right)\] we write \[\left( {1 + \tan A.\tan B} \right)\], that is just the opposite. Another point to note is that we have to take care of how the angles are expressed. Here, in this question all the angles are in degrees so we haven’t converted any of them, but in another question some angles might be given in degrees and some angles in radians. So, there we can convert all the angles into degrees or we can convert to radians.