Question

Find the value of $\tan {15^0} + \tan {75^0}$.
$
  \left( A \right){\text{. }}2\sqrt 3 \\
  \left( B \right).{\text{ }}2 \\
  \left( C \right).{\text{ }}2 - \sqrt 3 \\
  \left( D \right).{\text{ }}4\sqrt 3 \\
  \left( E \right).{\text{ }}4 \\
$

Answer 1

Hint- Try to make the expression of the form $\dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$ which is equal to$\sin 2\theta $.

Given: $\tan {15^0} + \tan {75^0}$
Rewriting above as:
$ = {\text{tan1}}{{\text{5}}^0} + \dfrac{1}{{\cot {{75}^0}}}{\text{ }}\left\{ {\because \tan \theta = \dfrac{1}{{\cot \theta }}} \right\}$
Now, using $\tan \theta = \cot \left( {{{90}^0} - \theta } \right)$ to convert $\cot $ into $\tan $, we get
$
   \Rightarrow \tan {15^0} + \dfrac{1}{{\cot \left( {{{90}^0} - {{15}^0}} \right)}} \\
   \Rightarrow \tan {15^0} + \dfrac{1}{{\tan {{15}^0}}}{\text{ }}\left\{ {\because \tan \theta = \cot \left( {{{90}^0} - \theta } \right)} \right\} \\
$
By cross-Multiplying, we get
$ \Rightarrow \dfrac{{{{\tan }^2}{{15}^0} + 1}}{{\tan {{15}^0}}}$
Multiply and divide by $2$, we get
$
   \Rightarrow \dfrac{{2\left[ {{{\tan }^2}{{15}^0} + 1} \right]}}{{2\tan {{15}^0}}} \\
   \Rightarrow \dfrac{2}{{\sin 2 \times {{15}^0}}}{\text{ }}\left\{ {\because \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = \sin 2\theta } \right\} \\
   \Rightarrow \dfrac{2}{{\sin {{30}^0}}} \\
$
Now, putting the value of $\sin {30^0} = \dfrac{1}{2}$, we get
$
   \Rightarrow \dfrac{2}{{\left( {\dfrac{1}{2}} \right)}} \\
   \Rightarrow 4 \\
$
$\therefore $ Correct option is $\left( E \right)$

Note- Whenever there are integer angles inside trigonometric functions whose value is not known to us, always try to convert it by using the $\left( {{{90}^0} - \theta } \right)$ form so as to make some trigonometric relations which makes problems much easier to solve.