# Prove that

Question:

Prove that

(i) $\frac{\cos 11^{\circ}+\sin 11^{\circ}}{\cos 11^{\circ}-\sin 11^{\circ}}=\tan 56^{\circ}$.

(ii) $\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}=\tan 54^{\circ}$

(ii) $\frac{\cos 8^{\circ}-\sin 8^{\circ}}{\cos 8^{\circ}+\sin 8^{\circ}}=\tan 37^{\circ}$

Solution:

(i) $\mathrm{LHS}=\frac{\cos 11^{\circ}+\sin 11^{\circ}}{\cos 11^{\circ}-\sin 11^{\circ}}$

$=\frac{\frac{\cos 11^{*}}{\cos 11^{*}}+\frac{\sin 11^{\circ}}{\cos 11^{\circ}}}{\frac{\cos 11^{*}}{\cos 11^{*}}-\frac{\sin 11^{\circ}}{\cos 11^{\circ}}}$ (Dividing numerator and denominator by $\cos 11^{\circ}$ )

$=\frac{1+\tan 11^{\circ}}{1-\tan 11^{\circ}}$

$=\frac{1+\tan 11^{\circ}}{1-1 \times \tan 11^{\circ}}$

$=\frac{\tan 45^{\circ}+\tan 11^{\circ}}{1-\tan 45^{\circ} \tan 11^{\circ}}$   $\left(\right.$ As $\left.\tan 45^{\circ}=1\right)$

$=\tan \left(45^{\circ}+11^{\circ}\right)$   $\left[\right.$ As $\left.\frac{\tan A+\tan B}{1-\tan A \tan B}=\tan (A+B)\right]$

$=\tan 56^{\circ}$

$=\operatorname{RHS}$

Hence proved.

(ii) $\mathrm{LHS}=\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}$

$=\frac{\frac{\cos 9^{\circ}}{\cos 9^{\circ}}+\frac{\sin 9^{\circ}}{\cos 9^{\circ}}}{\frac{\cos 9^{\circ}}{\cos 9^{\circ}}-\frac{\sin 9^{\circ}}{\cos 9^{\circ}}}$ (Dividing the numerator and denominator by $\cos 9$ )

$=\frac{1+\tan 9^{\circ}}{1-\tan 9^{\circ}}$

$=\frac{1+\tan 9^{\circ}}{1+1 \times \tan 9^{\circ}}$

$=\frac{\tan 45^{\circ}+\tan 9^{\circ}}{1-\tan 45^{\circ} \times \tan 9^{\circ}}$   $\left(\right.$ As $\left.\tan 45^{\circ}=1\right)$

$=\tan \left(45^{\circ}+9^{\circ}\right)$ $\left[A s \frac{\tan A+\tan B}{1-\tan A \tan B}=\tan (A+B)\right]$

$=\tan 54^{\circ}$

= RHS

Hence proved.

(iii) $\mathrm{LHS}=\frac{\cos 8^{\circ}-\sin 8^{\circ}}{\cos 8^{\circ}+\sin 8^{\circ}}$

$=\frac{\frac{\cos 8^{\circ}}{\cos 8^{\circ}}-\frac{\sin 8^{\circ}}{\cos 8^{\circ}}}{\frac{\cos 8}{\cos 8}+\frac{\sin 8}{\cos 8}}$ (Dividing numeraor and denominator by $\cos 8^{\circ}$ )

$=\frac{1-\tan 8^{\circ}}{1+\tan 8^{\circ}}$

$=\frac{1-\tan 8^{\circ}}{1+1 \times \tan 8^{\circ}}$

$=\frac{\tan 45^{\circ}-\tan 8^{\circ}}{1+\tan 45^{\circ} \tan 8^{\circ}} \quad\left(\right.$ As $\left.\tan 45^{\circ}=1\right)$

$=\tan \left(45^{\circ}-8^{\circ}\right)$ $\left[\right.$ As $\left.\frac{\tan A-\tan B}{1+\tan A \tan B}=\tan (A+B)\right]$

$=\tan 37^{\circ}$

= RHS

Hence proved.