Question

# The value of $\tan {81^0} - \tan {63^0} - \tan {27^0} + \tan {9^0}$is$\left( A \right).{\text{ 0}} \\ \left( B \right).{\text{ 2}} \\ \left( C \right).{\text{ 3}} \\ \left( D \right).{\text{ 4}} \\$

Verified
161.7k+ views
Hint: Try to write $\tan \theta$ in terms of $\cot \theta$, then use trigonometric formulas.

Given:$\tan {81^0} - \tan {63^0} - \tan {27^0} + \tan {9^0}$
Rewriting above as:
$\Rightarrow \left( {\tan {9^0} + \tan {{81}^0}} \right) - \left( {\tan {{27}^0} + \tan {{63}^0}} \right) \\ \Rightarrow \left( {\tan {9^0} + \tan \left( {{{90}^0} - {9^0}} \right)} \right) - \left( {\tan {{27}^0} + \tan \left( {{{90}^0} - {{27}^0}} \right)} \right){\text{ }} \ldots \left( A \right) \\$
We know that, $\tan \left( {{{90}^0} - \theta } \right) = \cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}{\text{ }} \ldots \left( 1 \right)$
Putting the value of $\tan \left( {{{90}^0} - \theta } \right)$from $\left( 1 \right)$in $\left( A \right)$, we get
$\Rightarrow \tan {9^0} + \cot {9^0} - \tan {27^0} - \cot {27^0}$
Now, since $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
$\therefore \left( {\dfrac{{\sin {9^0}}}{{\cos {9^0}}} + \dfrac{{\cos {9^0}}}{{\sin {9^0}}}} \right) - \left( {\dfrac{{\sin {{27}^0}}}{{\cos {{27}^0}}} + \dfrac{{\cos {{27}^0}}}{{\sin {{27}^0}}}} \right) \\ \Rightarrow \left( {\dfrac{{{{\sin }^2}{9^0} + {{\cos }^2}{9^0}}}{{\sin {9^0}\cos {9^0}}}} \right) - \left( {\dfrac{{{{\sin }^2}{{27}^0} + {{\cos }^2}{{27}^0}}}{{\sin {{27}^0}\cos {{27}^0}}}} \right) \\$
Using the property ${\sin ^2}\theta + {\cos ^2}\theta = 1$ in above equation, we get
$\Rightarrow \left( {\dfrac{1}{{\sin {9^0}\cos {9^0}}}} \right) - \left( {\dfrac{1}{{\sin {{27}^0}\cos {{27}^0}}}} \right)$
Also, we know that $\sin x\cos x = \dfrac{{\sin 2x}}{2}$
$\therefore \left( {\dfrac{2}{{\sin {{18}^0}}}} \right) - \left( {\dfrac{2}{{\sin {{54}^0}}}} \right) \\ \Rightarrow 2\left( {\dfrac{{\sin {{54}^0} - \sin {{18}^0}}}{{\sin {{18}^0}\sin {{54}^0}}}} \right) \\$
Now, by using $\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$ in above equation, we get
$\Rightarrow 2\left( {\dfrac{{2\cos \left( {\dfrac{{54 + 18}}{2}} \right)\sin \left( {\dfrac{{54 - 18}}{2}} \right)}}{{\sin {{18}^0}\sin \left( {{{90}^0} - {{36}^0}} \right)}}} \right)$
using $\sin \left( {{{90}^0} - \theta } \right) = \cos \theta$in the above equation, we get
$\Rightarrow 2\left( {\dfrac{{2\cos \left( {\dfrac{{72}}{2}} \right)\sin \left( {\dfrac{{36}}{2}} \right)}}{{\sin {{18}^0}\cos {{36}^0}}}} \right) \\ \Rightarrow 2\left( {\dfrac{{2\cos \left( {{{36}^0}} \right)\sin \left( {{{18}^0}} \right)}}{{\sin {{18}^0}\cos {{36}^0}}}} \right) \\ \Rightarrow 4\left( {\dfrac{{\sin {{18}^0}\cos {{36}^0}}}{{\sin {{18}^0}\cos {{36}^0}}}} \right) \\ \Rightarrow 4 \\$
$\therefore$the correct option is $\left( D \right)$.

Note: Whenever there are integer angles inside trigonometric functions whose values are not known to us, always try to convert them by using ${90^0} - \theta$ in order to make calculation easier.