Question

The value of $\tan {9^ \circ } - \tan {27^ \circ } - \tan {63^ \circ } + \tan {81^ \circ }$ is equal to
(A) -1
(B) 0
(C) 1
(D) 4

Answer 1

Hint : To find the value of a given expression we will rearrange the expression. Then we will use the trigonometric identity to rewrite the expression in the form $\cot$. Then we will again rewrite the expression in terms of sine and cosine. By again using the trigonometric identity we resolve the problem in a simpler form of sine and cosine. To convert the equation in the form of $\sin 2\theta$ we will multiply and divide the expression by 2. Then we will use the predefined formulas to get the answer.
Formula used:
We are using following trigonometric formulas:
1.$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
2.${\mathop{\rm Sin}\nolimits} 2\theta = 2\sin \theta \cos \theta$
3.${\cos ^2}\theta + {\sin ^2}\theta = 1$
4. $\tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta$

Complete step-by-step answer :
We have given expressions as$\tan {9^ \circ } - \tan {27^ \circ } - \tan {63^ \circ } + \tan {81^ \circ }$.
We will rearrange the given expression as:
$= \tan {9^ \circ } + \tan {81^ \circ } - \tan {27^ \circ } - \tan {63^ \circ }$
Then we will substitute ${81^ \circ }$ by $\left( {{{90}^ \circ } - {9^ \circ }} \right)$ and ${63^ \circ }$ by $\left( {{{90}^ \circ } - {{27}^ \circ }} \right)$. We will get,
$= \tan {9^ \circ } + \tan \left( {{{90}^ \circ } - {9^ \circ }} \right) - \tan {27^ \circ } - \tan \left( {{{90}^ \circ } - {{27}^ \circ }} \right)$
We know that $\tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta$. Hence using this expression in the above expression as:
$= \tan {9^ \circ } + \cot {9^ \circ } - \tan {27^ \circ } + \cot {27^ \circ }$
We know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ . hence using these expression, we can rewrite the above expression as:

= \dfrac{{\sin {9^ \circ }}}{{\cos {9^ \circ }}} + \dfrac{{\cos {9^ \circ }}}{{\sin {9^ \circ }}} - \dfrac{{\sin {{27}^ \circ }}}{{\cos {{27}^ \circ }}} - \dfrac{{\cos {{27}^ \circ }}}{{\sin {{27}^ \circ }}}\\\ = \dfrac{{{{\sin }^2}{9^ \circ } + {{\cos }^2}{9^ \circ }}}{{\sin {9^ \circ }\cos {9^ \circ }}} - \dfrac{{{{\sin }^2}{{27}^ \circ } + {{\cos }^2}{{27}^ \circ }}}{{\sin {{27}^ \circ }\cos {{27}^ \circ }}} \end{array}$$ We will use trigonometric identity ${\cos ^2}\theta + {\sin ^2}\theta = 1$ to rewrite the above expression: $$ = \dfrac{1}{{\sin {9^ \circ }\cos {9^ \circ }}} - \dfrac{1}{{\sin {{27}^ \circ }\cos {{27}^ \circ }}}$$ We multiply and divide the above expression by 2, we will get $$ = \dfrac{2}{{2\sin {9^ \circ }\cos {9^ \circ }}} - \dfrac{2}{{2\sin {{27}^ \circ }\cos {{27}^ \circ }}}$$ Now we know that ${\mathop{\rm Sin}\nolimits} 2\theta = 2\sin \theta \cos \theta $ , using this expression we will get $$ = \dfrac{2}{{\sin {{18}^ \circ }}} - \dfrac{2}{{\sin {{54}^ \circ }}}$$ We will substitute $\dfrac{{\sqrt 5 - 1}}{4}$ for $\sin {18^ \circ }$ and $\dfrac{{\sqrt 5 + 1}}{4}$ for $\sin {54^ \circ }$ in the above expression. $$\begin{array}{l} = \dfrac{{2 \times 4}}{{\sqrt 5 - 1}} - \dfrac{{2 \times 4}}{{\sqrt 5 + 1}}\\\ = 4 \end{array}$$ Hence the value of $\tan {9^ \circ } - \tan {27^ \circ } - \tan {63^ \circ } + \tan {81^ \circ }$ is equal to 4 and option (4) is correct. **So, the correct answer is “Option D”.** **Note** : In this question we are using predefined formulas and identities and hence we should have prior knowledge about the identities. The expression$\dfrac{{\sqrt 5 - 1}}{4}$ for $\sin {18^ \circ }$ and $\dfrac{{\sqrt 5 + 1}}{4}$ for $\sin {54^ \circ }$is derived in the trigonometry and hence we need to remember its value. Since, deduction of these values is a separate question.