Question

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

• A. $0$
• B. $2$
• C. $1$
• D. $4$

Answer 1

The correct option is D $4$

$=\tan 9^o-\tan {27}^o-\tan (90-27{)}^o+\tan (90-9{)}^o$

$=\tan 9^o-tan{27}^o-\cot {27}^o+\cot 9^o$

$=(\tan 9^o+\cot 9^o)(\tan {27}^o+\cot {27}^o)$

$=\frac{\sin 9^o}{\cos 9^o}+\frac{\cos 9^o}{\sin 9^o}-[\frac{\sin {27}^o}{\cos {27}^o}+\frac{\cos {27}^o}{\sin {27}^o}]$

$=\frac{si{n}^2{9}^o+{\cos }^2{9}^o}{\cos 9^o\sin 9^o}-\left[\frac{si{n}^2{27}^o+{\cos }^2{27}^o}{\cos {27}^o\sin {27}^o}\right]$

$=\frac{1}{\cos 9^o\sin 9^o}-\frac{1}{\cos {27}^o\sin {27}^o}$

$=\frac{2}{2\cos 9^o\sin 9^o}-\frac{2}{2\cos {27}^o\sin {27}^o}$

$=\frac{2}{\sin {18}^o}-\frac{2}{\sin {54}^o}$

$=2\left[\frac{\sin {54}^o-\sin {18}^o}{\sin {18}^o\sin {54}^o}\right]$

$=2\left[\frac{2\sin {36}^o\sin {18}^o}{\sin {18}^o\sin {54}^o}\right]$

$=4\frac{\cos {36}^o\sin {18}^o}{\sin {18}^o\cos {36}^o}$

$=4$