Question

evaluate: Sin 60°/cos^2 45°- cot30°+ 15 cos 90°

Answer 1

Hope this will help u.......

Answer 2

$\dfrac{\sin 60}{\cos ^{2}45} -\cot 30+15\cos 90=0$

Step-by-step explanation:

We have,

$\dfrac{\sin 60}{\cos ^{2}45} -\cot 30+15\cos 90$

Evaluate, $\dfrac{\sin 60}{\cos ^{2}45} -\cot 30+15\cos 90$ = ?

∴ $\dfrac{\sin 60}{\cos ^{2}45} -\cot 30+15\cos 90$

$=\dfrac{\dfrac{\sqrt{3}}{2} }{(\dfrac{1}{\sqrt{2}})^{2} } -\sqrt{3} +15(0)$

[ ∵ $\sin 60=\dfrac{\sqrt{3}}{2} ,\cos 45=\dfrac{1}{\sqrt{2}} ,\cot 30=\sqrt{3} and \cos 90=1$]

$=\dfrac{\dfrac{\sqrt{3}}{2} }{\dfrac{1}{2}} -\sqrt{3} +0$

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

= 0

Hence, $\dfrac{\sin 60}{\cos ^{2}45} -\cot 30+15\cos 90=0$