Question

Prove that $\frac{\tan 69°+\tan 66°}{1-\tan 69°\tan 66°}=-1$.

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{\tan 69°+\tan 66°}{1-\tan 69°\tan 66°} \\ =\tan \left(69°+66°\right)\left[\text{U}\mathrm{sing}\mathrm{the}\mathrm{formula}\frac{\tan A+\tan B}{1-\tan A\tan B}=\tan \left(A+B\right)\right] \\ =\tan 135° \\ =\tan \left(180°-45°\right) \\ =-\tan 45°\left[\tan \left(180-A\right)=-\tan A\right] \\ =-1 \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$