Question

Solve:
$\frac{\sin {60}^o}{{\cos }^2{45}^o}-3\tan {30}^o+5\cos {90}^o$

Answer 1

Now,

$\frac{\sin {60}^o}{{\cos }^2{45}^o}-3\tan {30}^o+5\cos {90}^o$

$=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}-3\times \frac{1}{\sqrt{3}}+5\times 0$ [ Using the values of $\sin$, $\cos$ and $\tan$ 's]

$=\sqrt{3}-\sqrt{3}+0$

$=0$.