Question:
Prove that $\frac{\tan 69^{\circ}+\tan 66^{\circ}}{1-\tan 69^{\circ} \tan 66^{\circ}}=-1$.
Solution:
$\mathrm{LHS}=\frac{\tan 69^{\circ}+\tan 66^{\circ}}{1-\tan 69^{\circ} \tan 66^{\circ}}$
$=\tan \left(69^{\circ}+66^{\circ}\right) \quad\left[\mathrm{U} \operatorname{sing}\right.$ the formula $\left.\frac{\tan A+\tan B}{1-\tan A \tan B}=\tan (A+B)\right]$
$=\tan 135^{\circ}$
$=\tan \left(180^{\circ}-45^{\circ}\right)$
$=-\tan 45^{\circ} \quad[\tan (180-A)=-\tan A]$
$=-1$
= RHS
Hence proved.