# 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

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Answer

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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.

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.

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