Question

Prove that $\frac{\tan {69}^o+\tan {66}^o}{1-\tan {69}^o\tan {66}^o}=-1$

Answer 1

Now,

$\frac{\cos 9^o+\sin 9^o}{\cos 9^o-\sin 9^o}$

$=\frac{1+\tan 9^o}{1-\tan 9^o}$ [ Dividing the numerator and the denominator of the above expression by $\cos 9^o$]

$=\frac{\tan {45}^o+\tan 9^o}{1-\tan {45}^o.\tan 9^o}$ [ Since $\tan {45}^o=1$]

$=\tan ({45}^o+9^o)$

$=\tan {54}^o$.