Question

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

Answer 1

$$\begin{gathered} 2\left(\frac{\cos 58°}{\sin 32°}\right)-\sqrt{3}\left(\frac{\cos 38°\mathrm{cosec}52°}{\tan 15°\tan 60°\tan 75°}\right) \\ =2\left(\frac{\cos \left(90°-32°\right)}{\sin 32°}\right)-\sqrt{3}\left(\frac{\cos \left(90°-52°\right)\mathrm{cosec}52°}{\tan 15°\tan 60°\tan 75°}\right) \\ =2\left(\frac{\sin 32°}{\sin 32°}\right)-\sqrt{3}\left(\frac{\sin 52°\mathrm{cosec}52°}{\tan 15°\tan 60°\tan 75°}\right)\left(\because \cos \left(90°-\theta \right)=\sin \theta \right) \\ =2\left(1\right)-\sqrt{3}\left(\frac{\sin 52°{\displaystyle \frac{1}{\sin 52°}}}{\tan 15°\tan 60°\tan 75°}\right)\left(\because \mathrm{cosec}\theta =\frac{1}{\sin \theta }\right) \\ =2-\sqrt{3}\left(\frac{1}{\tan 15°\tan 60°\tan 75°}\right) \\ =2-\sqrt{3}\left(\frac{1}{\tan 15°\tan 60°\tan \left(90°-15°\right)}\right) \\ =2-\sqrt{3}\left(\frac{1}{\tan 15°\tan 60°\cot 15°}\right)\left(\because \tan \left(90°-\theta \right)=\cot \theta \right) \\ =2-\sqrt{3}\left(\frac{1}{\tan 15°\tan 60°{\displaystyle \frac{1}{\tan 15°}}}\right)\left(\because \cot \theta =\frac{1}{\tan \theta }\right) \\ =2-\sqrt{3}\left(\frac{1}{\tan 60°}\right) \\ =2-\sqrt{3}\left(\frac{1}{\sqrt{3}}\right)\left(\because \tan 60°=\sqrt{3}\right) \\ =2-1 \\ =1 \\ \mathrm{Hence},2\left(\frac{\cos 58°}{\sin 32°}\right)-\sqrt{3}\left(\frac{\cos 38°\mathrm{cosec}52°}{\tan 15°\tan 60°\tan 75°}\right)=1. \end{gathered}$$