Question

Prove that:

(i) $\frac{\cos 9°+\sin 9°}{\cos 9°-\sin 9°}=\tan 54°$
(ii) $\frac{\cos 8°-\sin 8°}{\cos 8°+\sin 8°}=\tan 37°$

Answer 1

$$\begin{gathered} \left(\text{i}\right)\mathrm{LHS}=\frac{\cos 9°+\sin 9°}{\cos 9°-\sin 9°} \\ =\frac{{\displaystyle \frac{\cos 9°}{\cos 9°}}+{\displaystyle \frac{\sin 9°}{\cos 9°}}}{{\displaystyle \frac{\cos 9°}{\cos 9°}}-{\displaystyle \frac{\sin 9°}{\cos 9°}}}\left(\mathrm{Dividing}\mathrm{the}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}\cos 9\right) \\ =\frac{1+\tan 9°}{1-\tan 9°} \\ =\frac{1+\tan 9°}{1+1\times \tan 9°} \\ =\frac{\tan 45°+\tan 9°}{1-\tan 45°\times \tan 9°}\left(\mathrm{As}\tan 45°=1\right) \\ =\tan \left(45°+9°\right)\left[As\frac{\tan A+\tan B}{1-\tan A\tan B}=\tan \left(A+B\right)\right] \\ =\tan 54° \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{LHS}=\frac{\cos 8°-\sin 8°}{\cos 8°+\sin 8°} \\ =\frac{{\displaystyle \frac{\cos 8°}{\cos 8°}}-{\displaystyle \frac{\sin 8°}{\cos 8°}}}{{\displaystyle \frac{\cos 8}{\cos 8}}+{\displaystyle \frac{\sin 8}{\cos 8}}}\left(\mathrm{Dividing}\mathrm{numeraor}\mathrm{and}\mathrm{denominator}\mathrm{by}\cos 8^{°}\right) \\ =\frac{1-\tan 8°}{1+\tan 8°} \\ =\frac{1-\tan 8°}{1+1\times \tan 8°} \\ =\frac{\tan 45°-\tan 8°}{1+\tan 45°\tan 8°}\left(\mathrm{As}\tan 45°=1\right) \\ =\tan \left(45°-8°\right)\left[\mathrm{As}\frac{\tan A-\tan B}{1+\tan A\tan B}=\tan \left(A+B\right)\right] \\ =\tan 37° \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$