Evaluate the following:frac{cos 45^{circ}}{se | vedclass

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Question

(A) Given expression: $\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$Substitute the trigonometric values: $\cos 45^{\circ} =

Vedclass · Accepted Answer

$\frac{3\sqrt{2}-\sqrt{6}}{8}$

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(A) Given expression: $\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$Substitute the trigonometric values: $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,$\sec 30^{\circ} = \frac{2}{\sqrt{3}}$,and $\operatorname{cosec} 30^{\circ} = 2$.$\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2} = \frac{\frac{1}{\sqrt{2}}}{\frac{2+2\sqrt{3}}{\sqrt{3}}}$$= \frac{\sqrt{3}}{\sqrt{2}(2+2\sqrt{3})} = \frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}$Rationalize the denominator by multiplying the numerator and denominator by $(2\sqrt{6}-2\sqrt{2})$:$= \frac{\sqrt{3}(2\sqrt{6}-2\sqrt{2})}{(2\sqrt{6}+2\sqrt{2})(2\sqrt{6}-2\sqrt{2})}$$= \frac{2\sqrt{18}-2\sqrt{6}}{(2\sqrt{6})^2-(2\sqrt{2})^2} = \frac{2(3\sqrt{2})-\sqrt{6}}{24-8} = \frac{6\sqrt{2}-2\sqrt{6}}{16}$$= \frac{2(3\sqrt{2}-\sqrt{6})}{16} = \frac{3\sqrt{2}-\sqrt{6}}{8}$

Evaluate the following:
$\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$

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Evaluate the following:$\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$