Question

sin45cos45cos60/sin60cos30tan45

Answer 1

Answer:

$\dfrac{Sin45.Cos45.Cos60}{Sin60.Cos30.tan45}$ = $\dfrac{1}{3}$

Step-by-step explanation:

We have-

$\dfrac{Sin45.Cos45.Cos60}{Sin60.Cos30.tan45}$

We know that-

$Sin45 = \dfrac{1}{\sqrt{2}}$

$Cos45 = \dfrac{1}{\sqrt{2}}$

$Cos60 = \dfrac{1}{2}$

$Sin60 = \dfrac{\sqrt{3} }{2}$

$Cos30 = \dfrac{\sqrt{3} }{2}$

tan45 = 1

So, $\dfrac{Sin45.Cos45.Cos60}{Sin60.Cos30.tan45}$

= $\dfrac{\dfrac{1}{\sqrt{2}} \times \dfrac{1}{\sqrt{2} } \times \dfrac{1}{2}   }{\dfrac{\sqrt{3} }{2} \times\dfrac{\sqrt{3} }{2} \times 1  }$

$= \dfrac{\dfrac{1}{2} \times\dfrac{1}{2}  }{\dfrac{3}{4} }$

$= \dfrac{1}{4} \times \dfrac{4}{3}$

$= \dfrac{1}{3}$

Hence, $\dfrac{Sin45.Cos45.Cos60}{Sin60.Cos30.tan45}$ = $\dfrac{1}{3}$