Question

(cos60° - sin45° ) ÷ (sec30°) By steps....​

Answer 1

Answer:

The answer is $\dfrac{\sqrt 3-\sqrt 6}{4}$

Step-by-step explanation:

                            Using the results :

        $\boxed{\cos60^\circ=\dfrac{1}{2}, \sin45^\circ=\dfrac{1}{\sqrt 2}=\dfrac{\sqrt 2}{2}, \sec 30^\circ=\dfrac{2}{\sqrt 3}}$

Thus the value of  

              $\dfrac{\cos60^\circ-\sin45^\circ}{\sec 30^\circ}$

               $=\dfrac{\frac{1}{2}-\frac{\sqrt 2}{2}}{\frac{2}{\sqrt 3}}$

                $=\dfrac{1-\sqrt 2}{2(2)}\times \sqrt 3=\dfrac{\sqrt 3-\sqrt 6}{4}$