Prove that :
(i) $\tan 20^{\circ} \tan 35^{\circ} \tan 45^{\circ} \tan 55^{\circ} \tan 70^{\circ}=1$
(ii) $\sin 48^{\circ} \sec 42^{\circ}+\cos 48^{\circ} \operatorname{cosec} 42^{\circ}=2$
(iii) $\frac{\sin 70^{\circ}}{\cos 20^{\circ}}+\frac{\operatorname{cosec} 20^{\circ}}{\sec 70^{\circ}}-2 \cos 70^{\circ} \operatorname{cosec} 20^{\circ}=0$
(iv) $\frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \operatorname{cosec} 31^{\circ}=2$
We are asked to find the value of $\tan 20^{\circ} \tan 35^{\circ} \tan 45^{\circ} \tan 55^{\circ} \tan 70^{\circ}$
(i) Therefore
$\tan 20^{\circ} \tan 35^{\circ} \tan 45^{\circ} \tan 55^{\circ} \tan 70^{\circ}$
$=\tan \left(90^{\circ}-70^{\circ}\right) \tan \left(90^{\circ}-55^{\circ}\right) \tan 45^{\circ} \tan 55^{\circ} \tan 70^{\circ}$
$=\cot 70^{\circ} \cot 55^{\circ} \tan 45^{\circ} \tan 55^{\circ} \tan 70^{\circ}$
$=\left(\tan 70^{\circ} \cot 70^{\circ}\right)\left(\tan 55^{\circ} \cot 55^{\circ}\right) \tan 45^{\circ}$
$=1 \times 1 \times 1$
$=1$
Proved
(ii) We will simplify the left hand side
$\sin 48^{\circ} \cdot \sec 48^{\circ}+\cos 48^{\circ} \cdot \operatorname{cosec} 42^{\circ}=\sin 48^{\circ} \cdot \sec \left(90^{\circ}-48^{\circ}\right)+\cos 48^{\circ} \cdot \operatorname{cosec}\left(90^{\circ}-48^{\circ}\right)$
$=\sin 48 \cdot \cos 48^{\circ}+\cos 48^{\circ} \cdot \sin 48^{\circ}$
$=1+1$
$=2$
Proved
(iii) We have, $\frac{\sin 70^{\circ}}{\cos 20^{\circ}}+\frac{\operatorname{cosec} 20^{\circ}}{\sec 70^{\circ}}-2 \cos 70^{\circ} \operatorname{cosec} 20^{\circ}=0$
So we will calculate left hand side
$\frac{\sin 70^{\circ}}{\cos 20^{\circ}}+\frac{\operatorname{cosec} 20^{\circ}}{\sec 70^{\circ}}-2 \cos 70^{\circ} \cdot \operatorname{cosec} 20^{\circ}=\frac{\sin 70^{\circ}}{\cos 20^{\circ}}+\frac{\cos 70^{\circ}}{\sin 20^{\circ}}-2 \cos 70^{\circ} \cdot \operatorname{cosec}\left(90^{\circ}-70^{\circ}\right)$
$=\frac{\sin \left(90^{\circ}-20^{\circ}\right)}{\cos 20^{\circ}}+\frac{\cos \left(90^{\circ}-20^{\circ}\right)}{\sin 20^{\circ}}-2 \cos 70^{\circ} \cdot \sec 70^{\circ}$
$=\frac{\cos 20^{\circ}}{\cos 20^{\circ}}+\frac{\sin 20^{\circ}}{\sin 20^{\circ}}-2 \times 1$
$=1+1-2$
$=2-2$
$=0$
Proved
(iv) We have $\frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \cdot \operatorname{cosec} 31^{\circ}=2$
We will simplify the left hand side
$\frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \cdot \operatorname{cosec} 31^{\prime}=\frac{\cos \left(90^{\circ}-10^{\circ}\right)}{\sin 10}+\cos 59^{\circ} \cdot \operatorname{cosec}\left(90^{\circ}-59^{\circ}\right)$
$=\frac{\sin 10^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \cdot \sec 59^{\circ}$
$=1+1$
$=2$
Proved