Courses
Courses for Kids
Free study material
Free LIVE classes
More

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

Last updated date: 20th Mar 2023
Total views: 307.8k
Views today: 3.86k
Verified
307.8k+ views
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.