Question

 The value of \[\;\tan 81^\circ - \tan 63^\circ - \tan 27^\circ + \tan 9^\circ \] is?
A.1
B.2
C.3
D.4

Answer 1

Hint: First we convert$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$, and then simplify after that use the formula
\[\;\sin A\cos B + \cos A\sin B = \sin \left( {A + B} \right)\] Apply this formula in numerator we get\[\sin 90^\circ \]which is equal to 1. After that we can use the formula$2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)$ in denominator then substituted the value of\[\cos {90^ \circ }\], $\cos 72^\circ = $$\left( {\dfrac{{\surd 5 - 1}}{4}} \right)$and$\cos 36^\circ = \left( {\dfrac{{\surd 5 + 1}}{4}} \right)$ then simplify it in order to get answer.
Complete step-by-step solution:
Given: \[\;\tan 81^\circ - \tan 63^\circ - \tan 27^\circ + \tan 9^\circ \]
\[ \Rightarrow \]\[\;\left( {\tan 81^\circ + \tan 9^\circ } \right)\]
Now we use $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$,
\[ \Rightarrow \]$\left( {\dfrac{{\sin 81^\circ }}{{\cos 81^\circ }} + \dfrac{{\sin 9^\circ }}{{\cos 9^\circ }}} \right)$ $\left( {\dfrac{{\sin 3^\circ }}{{\cos 63^\circ }} + \dfrac{{\sin 27^\circ }}{{\cos 27^\circ }}} \right)$
On further simplification we get,
\[ \Rightarrow \]$\left( {\dfrac{{\sin 81^\circ \cos 9^\circ + \cos 81^\circ \sin 9^\circ }}{{\cos 81^\circ \cos 9^\circ }}} \right) - \left( {\dfrac{{\sin 63^\circ \cos 27^\circ + \cos 63^\circ \sin 27^\circ }}{{\cos 63^\circ \cos 27^\circ }}} \right)$
Now we use the formula\[\sin (A + B) = \sin A\cos B + \cos A\sin B\]and simplify further,
\[ \Rightarrow \]\[\left( {\dfrac{{\sin \left( {81^\circ + 9^\circ } \right)}}{{\cos 81^\circ \cos 9^\circ }}} \right) - \left( {\dfrac{{\sin \left( {63^\circ + 27^\circ } \right)}}{{\cos 63^\circ \cos 27^\circ }}} \right)\]
\[ \Rightarrow \left( {\dfrac{{\sin 90^\circ }}{{\cos 81^\circ \cos 9^\circ }}} \right) - \left( {\dfrac{{\sin 90^\circ }}{{\cos 63^\circ \cos 27^\circ }}} \right)\]
Now we substitute the value of \[\sin {90^ \circ } = 1\], we get,
\[ \Rightarrow \left( {\dfrac{2}{{2\cos 81^\circ \cos 9^\circ }}} \right) - \left( {\dfrac{2}{{2\cos 63^\circ \cos 27^\circ }}} \right)\]
Now multiply and divide by 2,
\[ \Rightarrow \left( {\dfrac{2}{{2\cos 81^\circ \cos 9^\circ }}} \right) - \left( {\dfrac{2}{{2\cos 63^\circ \cos 27^\circ }}} \right)\]
Now we use the formula \[\cos \left( {A + B} \right) + \cos \left( {A - B} \right) = 2\cos A\cos B\]and on further simplification we get,
\[ \Rightarrow \left( {\dfrac{2}{{\cos \left( {81^\circ + 9^\circ } \right) + \cos \left( {81^\circ - 9^\circ } \right)}}} \right) - \left( {\dfrac{2}{{\cos \left( {63^\circ + 27^\circ } \right) + \cos \left( {63^\circ - 27^\circ } \right)}}} \right)\]
\[ \Rightarrow \left( {\dfrac{2}{{\cos 90^\circ + \cos 72^\circ }}} \right) - \left( {\dfrac{2}{{\cos 90^\circ + \cos 36^\circ }}} \right)\]
Now we substitute the value of \[\cos {90^ \circ } = 0\], we get,
\[ \Rightarrow \left( {\dfrac{2}{{\cos 72^\circ }}} \right) - \left( {\dfrac{2}{{\cos 36^\circ }}} \right)\]
Now we substitute the value of $\cos 36^\circ = \left( {\dfrac{{\surd 5 + 1}}{4}} \right)$ and $\cos 72^\circ = \left( {\dfrac{{\surd 5 - 1}}{4}} \right)$, we get,
\[ \Rightarrow \left( {\dfrac{{2 \times 4}}{{\surd 5 - 1}}} \right) - \left( {\dfrac{{2 \times 4}}{{\surd 5 + 1}}} \right)\]
On further simplification we get,
$ \Rightarrow \left( {8\left( {\dfrac{1}{{\surd 5 - 1}} - \dfrac{1}{{\surd 5 + 1}}} \right)} \right)$
$ \Rightarrow \left( {8\left( {\dfrac{{\surd 5 + 1 - \surd 5 + 1}}{{\left( {\surd 5 - 1} \right)\left( {\surd 5 + 1} \right)}}} \right)} \right)$
$ \Rightarrow \left( {8\left( {\dfrac{2}{4}} \right)} \right)$
$ \Rightarrow 4$
 Hence, Option D. 4 is the correct answer.
Note: In this type of simplification question first try to convert the tan function into sine and cosine so it will be convenient to apply the required formula in order to get the value of expression. Also remember some standard values such as \[\cos {90^ \circ } = 0\], \[\sin {90^ \circ } = 1\], \[\cos {0^ \circ } = 1\], \[\sin {0^ \circ } = 0\],\[\cos {180^ \circ } = - 1\], \[sin{180^ \circ } = 0\]