Answer
Verified
500.4k+ 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.
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.
Recently Updated Pages
Write the IUPAC name of the given compound class 11 chemistry CBSE
Write the IUPAC name of the given compound class 11 chemistry CBSE
Write the IUPAC name of the given compound class 11 chemistry CBSE
Write the IUPAC name of the given compound class 11 chemistry CBSE
Write the IUPAC name of the given compound class 11 chemistry CBSE
Write the IUPAC name of the given compound class 11 chemistry CBSE
Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
In the tincture of iodine which is solute and solv class 11 chemistry CBSE
On which part of the tongue most of the taste gets class 11 biology CBSE
State and prove Bernoullis theorem class 11 physics CBSE
Who is the leader of the Lok Sabha A Chief Minister class 11 social science CBSE