Question

tan9 - tan27 - tan63 + tan 81 = ?

Answer 1

hmmm they're all factors of 9....

Answer 2

NOTE : 81+9=90
63+27=90

Answer 3

ahh yes..that...

Answer 4

that wont help...@hartnn i think...

Answer 5

really ?
tan 81 = cot 9 = 1/ tan 9
now??

Answer 6

lol....Go on..u will.....see....

Answer 7

hmmm.....sin 18 and sin 54...

Answer 8

The answer should be 4

Answer 9

u need sin 18 = \(\sqrt5-1\) else u cannot solve it.

Answer 10

Cant we Do..this... With out Knwing sin 18....i mean,,,,any other method

Answer 11

tan 9=sin9/cos9=cos81/sin81=cot81
tan 27=sin27/cos27=cos63/sin63=cot27
so!
tan9+cot81-tan63-cot27=?
remember the relation b/n tan and cot of complementary angles
tan a=cot 90-a

Answer 12

\[\tan9 - \tan 27 - \tan63 - \tan81 \]
\[ \tan9 + \tan81 - \tan27 - \tan63\]
\[\frac{\sin9}{\cos9}+\frac{\sin 81}{\cos 81}-\frac{\sin 27}{\cos27}-\frac{\sin63}{\cos63}\]
\[ \frac{\sin90}{ \cos81\cos9} -\frac {\sin90}{ \cos63\cos27}\]
\[ \frac{1}{\sin9\cos9} - \frac{1}{\sin27\cos27}\]
\[ \frac{2}{\sin18} - \frac{2}{\sin54}\]
\[ 2\frac{\sin54 - \sin18}{\sin18\sin54}\]
\[ \frac{4\cos36\sin18} {\sin18\cos36}=4\]