Question

Evaluate:
$2\left(\frac{\cos 58}{\sin 32}\right)-\sqrt{3}\left(\frac{\cos 38\cos ec52}{\tan 15\tan 60\tan 75}\right)$

Answer 1

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

$2\left(\frac{cos{58}^{\circ }}{sin(90-58)}\right)-\sqrt{3}\left(\frac{cos{38}^{\circ }cosec(90-38)}{tan(90-75{)}^{\circ }tan{60}^{\circ }tan{75}^{\circ }}\right)$

As we know,

$sin(90-\theta )=cos\theta$

$cosec(90-\theta )=sec\theta$

$tan(90-\theta )=cot\theta$

$2\left(\frac{cos{58}^{\circ }}{cos{58}^{\circ }}\right)=\sqrt{3}\left(\frac{cos{38}^{\circ }.sec{38}^{\circ }}{tan{60}^{\circ }.tan75cot75}\right)$

Also,

$sec\theta =\frac{1}{cos\theta }$ and $cot\theta =\frac{1}{tan\theta }$

$2-\sqrt{3}\left(\frac{cos{38}^{\circ }.1/cos{38}^{\circ }}{tan{60}^{\circ }.tan{75}^{\circ }.1/tan{75}^{\circ }}\right)$

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

putting value of $tan{60}^{\circ }=\sqrt{3}$

$2-\sqrt{3}\left(\frac{1}{\sqrt{3}}\right)=2-1=1$