# Solve this

Question:

$2\left(\frac{\cos 58^{\circ}}{\sin 32^{\circ}}\right)-\sqrt{3}\left(\frac{\cos 38^{\circ} \operatorname{cosec} 52^{\circ}}{\tan 15^{\circ} \tan 60^{\circ} \tan 75^{\circ}}\right)$

Solution:

$2\left(\frac{\cos 58^{\circ}}{\sin 32^{\circ}}\right)-\sqrt{3}\left(\frac{\cos 38^{\circ} \operatorname{cosec} 52^{\circ}}{\tan 15^{\circ} \tan 60^{\circ} \tan 75^{\circ}}\right)$

$=2\left(\frac{\cos \left(90^{\circ}-32^{\circ}\right)}{\sin 32^{\circ}}\right)-\sqrt{3}\left(\frac{\cos \left(90^{\circ}-52^{\circ}\right) \operatorname{cosec} 52^{\circ}}{\tan 15^{\circ} \tan 60^{\circ} \tan 75^{\circ}}\right)$

$=2\left(\frac{\sin 32^{\circ}}{\sin 32^{\circ}}\right)-\sqrt{3}\left(\frac{\sin 52^{\circ} \operatorname{cosec} 52^{\circ}}{\tan 15^{\circ} \tan 60^{\circ} \tan 75^{\circ}}\right) \quad\left(\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right)$

$=2(1)-\sqrt{3}\left(\frac{\sin 52^{\circ} \frac{1}{\sin 52^{2}}}{\tan 15^{\circ} \tan 60^{\circ} \tan 75^{\circ}}\right) \quad\left(\because \operatorname{cosec} \theta=\frac{1}{\sin \theta}\right)$

$=2-\sqrt{3}\left(\frac{1}{\tan 15^{\circ} \tan 60^{\circ} \tan 75^{\circ}}\right)$

$=2-\sqrt{3}\left(\frac{1}{\tan 15^{\circ} \tan 60^{\circ} \tan \left(90^{\circ}-15^{\circ}\right)}\right)$

$=2-\sqrt{3}\left(\frac{1}{\tan 15^{\circ} \tan 60^{\circ} \cot 15^{\circ}}\right) \quad\left(\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right)$

$=2-\sqrt{3}\left(\frac{1}{\tan 15^{\circ} \tan 60^{\circ} \frac{1}{\tan 15^{-}}}\right) \quad\left(\because \cot \theta=\frac{1}{\tan \theta}\right)$

$=2-\sqrt{3}\left(\frac{1}{\tan 60^{\circ}}\right)$

$=2-\sqrt{3}\left(\frac{1}{\sqrt{3}}\right) \quad\left(\because \tan 60^{\circ}=\sqrt{3}\right)$

$=2-1$

$=1$

Hence, $2\left(\frac{\cos 58^{\circ}}{\sin 32^{\circ}}\right)-\sqrt{3}\left(\frac{\cos 38^{\circ} \operatorname{cosec} 52^{\circ}}{\tan 15^{\circ} \tan 60^{\circ} \tan 75^{\circ}}\right)=1$.