# Prove that :Prove that :

Question:

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) $\frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \operatorname{cosec} 31^{\circ}=2$

Solution:

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