Question

Find the value of $\tan {{9}^{\circ }}-\tan {{27}^{\circ }}-\tan {{63}^{\circ }}+\tan {{81}^{\circ }}$.

Answer 1

We convert the tangents of angles $\tan {{81}^{\circ }},\tan {{63}^{\circ }}$ to their corresponding co-tangents using the complimentary angle relation $\tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta$. We convert the tangent and cotangent of the angles to corresponding using the identity $\tan \theta =\dfrac{\sin \theta }{\cos \theta },\cot \theta =\dfrac{\cos \theta }{\sin \theta }$. We simplify further using the sine double angle formula $\sin 2\theta =2\sin \theta \cos \theta$ and difference of sine of angles formula $\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$.$$$$

Complete step-by-step solution:
We know if there are two complementary angles say $\theta$ and ${{90}^{\circ }}-\theta$ then the relation between tangent and cotangent are given by
$\tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta$
The relation between sine and sine are given by
$\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta$
We can convert tangent and cotangent of an angle $\theta$ using the identity
$\tan \theta =\dfrac{\sin \theta }{\cos \theta },\cot \theta =\dfrac{\cos \theta }{\sin \theta }$
The difference of sine of angles formula for some angles $C,D$ are given by,
$\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$
We are given the expression in tangent of angles from the question as
$\tan {{9}^{\circ }}-\tan {{27}^{\circ }}-\tan {{63}^{\circ }}+\tan {{81}^{\circ }}$
We observe that there are two pairs of complementary angles ${{9}^{\circ }},{{81}^{\circ }}$ and ${{27}^{\circ }},{{81}^{\circ }}$. Let us write them close to each other.

& \tan {{9}^{\circ }}-\tan {{27}^{\circ }}-\tan {{63}^{\circ }}+\tan {{81}^{\circ }}=\tan {{9}^{\circ }}+\tan {{81}^{\circ }}-\tan {{27}^{\circ }}-\tan {{63}^{\circ }} \\\ & =\tan {{9}^{\circ }}+\tan {{81}^{\circ }}-\left( \tan {{27}^{\circ }}+\tan {{63}^{\circ }} \right) \\\ \end{aligned}$$ We use the relation of complementary angles of tangent and cotangent two complementary angles in the above step for $\theta ={{9}^{\circ }},{{27}^{\circ }}$ and have $$\begin{aligned} & =\tan {{9}^{\circ }}+\tan {{\left( 90-9 \right)}^{\circ }}-\left( \tan {{27}^{\circ }}+\tan {{\left( 90-27 \right)}^{\circ }} \right) \\\ & =\tan {{9}^{\circ }}+\cot {{9}^{\circ }}-\left( \tan {{27}^{\circ }}+\cot {{27}^{\circ }} \right) \\\ \end{aligned}$$ Let us convert the tangent and cotangent of the angles in the above step to sine and cosines. We have $$\begin{aligned} &=\dfrac{\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}+\dfrac{\cos {{9}^{\circ }}}{\sin {{9}^{\circ }}}-\left( \dfrac{\sin {{27}^{\circ }}}{\cos {{27}^{\circ }}}+\dfrac{\cos {{27}^{\circ }}}{\sin {{27}^{\circ }}} \right) \\\ &=\dfrac{{{\sin }^{2}}{{9}^{\circ }}+{{\cos }^{2}}{{9}^{\circ }}}{\sin {{9}^{\circ }}\cos {{9}^{\circ }}}-\left( \dfrac{{{\sin }^{2}}{{27}^{\circ }}+{{\cos }^{2}}{{27}^{\circ }}}{\sin {{27}^{\circ }}\cos {{27}^{\circ }}} \right) \\\ \end{aligned}$$ Let use the algebraic identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ for any acute angle $\theta $ and have $$=\dfrac{1}{\sin {{9}^{\circ }}\cos {{9}^{\circ }}}-\dfrac{1}{\sin {{27}^{\circ }}\cos {{27}^{\circ }}}$$ Let us multiply 2 in numerator and denominator of the each term at left hand side of the above step and have, $$=\dfrac{2}{2\sin {{9}^{\circ }}\cos {{9}^{\circ }}}-\dfrac{2}{2\sin {{27}^{\circ }}\cos {{27}^{\circ }}}$$ We use the sine double angle formula for $\theta ={{9}^{\circ }}$ in the first term and for $\theta ={{27}^{\circ }}$ in the second term. We have, $$\begin{aligned} & =\dfrac{2}{\sin \left( 2\times {{9}^{\circ }} \right)}-\dfrac{2}{\sin \left( 2\times {{27}^{\circ }} \right)} \\\ & =2\left( \dfrac{1}{\sin {{18}^{\circ }}}-\dfrac{1}{\sin {{54}^{\circ }}} \right) \\\ & =2\left( \dfrac{\sin {{54}^{\circ }}-\sin {{18}^{\circ }}}{\sin {{18}^{\circ }}\sin {{54}^{\circ }}} \right) \\\ \end{aligned}$$ We use the complimentary angle relation between sine and cosine in the denominator to have $$\begin{aligned} & =2\left( \dfrac{\sin {{54}^{\circ }}-\sin {{18}^{\circ }}}{\sin {{18}^{\circ }}\sin {{\left( 90-36 \right)}^{\circ }}} \right) \\\ & =2\left( \dfrac{\sin {{54}^{\circ }}-\sin {{18}^{\circ }}}{\sin {{18}^{\circ }}\cos {{36}^{\circ }}} \right) \\\ \end{aligned}$$ We use the difference of sine of angles formula for $C={{54}^{\circ }},D={{18}^{\circ }}$in the numerator to have, $$\begin{aligned} &=2\left( \dfrac{2\cos \left( \dfrac{{{54}^{\circ }}+{{18}^{\circ }}}{2} \right)\sin \left( \dfrac{{{54}^{\circ }}-{{18}^{\circ }}}{2} \right)}{\sin {{18}^{\circ }}\cos {{36}^{\circ }}} \right) \\\ & =2\left( \dfrac{2\cos {{36}^{\circ }}\sin {{18}^{\circ }}}{\sin {{18}^{\circ }}\cos {{36}^{\circ }}} \right) \\\ \end{aligned}$$ We divide $\sin {{18}^{\circ }}\cos {{36}^{\circ }}$ in the numerator and denominator since $\sin {{18}^{\circ }}\cos {{36}^{\circ }}\ne 0 $. We have, $$\tan {{9}^{\circ }}-\tan {{27}^{\circ }}-\tan {{63}^{\circ }}+\tan {{81}^{\circ }}=2\times 2=4$$ **Note:** We can alternatively begin by converting $\cot \theta =\dfrac{1}{\tan \theta }$ and then proceed to use ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta .$ We can also begin with tangent sum of angles formula $\tan \left( A+B \right)=\dfrac{\tan A-\tan B}{1-\tan A\cdot \tan B}$ and then use the value of $\sin {{18}^{\circ }}=\dfrac{4}{\sqrt{5}-1}$ and $\sin {{54}^{\circ }}=\dfrac{4}{\sqrt{5}+1}$ which we can obtain taking sine of the equation $2A+3A={{90}^{\circ }}$ for $A={{18}^{\circ }}$ ad the using $\sin 3A=3\sin A-4{{\sin }^{3}}A$.