Solve 4/cot^{230^{circ}}+1/sin^260^circ-cos^245^circ

$$\frac { 4 } { \cot ^ { 2 } 30 ^ { \circ } } + \frac { 1 } { \sin ^ { 2 } 60 ^ { \circ } } - \cos ^ { 2 } 45 ^ { \circ }$$

Evaluate

$$\frac{13}{6}\approx 2.166666667$$

Solution Steps

Get the value of \cot(30) from trigonometric values table.

$$\frac{4}{\left(\sqrt{3}\right)^{2}}+\frac{1}{\left(\sin(60)\right)^{2}}-\left(\cos(45)\right)^{2}$$

The square of \sqrt{3} is 3.

$$\frac{4}{3}+\frac{1}{\left(\sin(60)\right)^{2}}-\left(\cos(45)\right)^{2}$$

Get the value of \sin(60) from trigonometric values table.

$$\frac{4}{3}+\frac{1}{\left(\frac{\sqrt{3}}{2}\right)^{2}}-\left(\cos(45)\right)^{2}$$

To raise \frac{\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{4}{3}+\frac{1}{\frac{\left(\sqrt{3}\right)^{2}}{2^{2}}}-\left(\cos(45)\right)^{2}$$

Divide 1 by \frac{\left(\sqrt{3}\right)^{2}}{2^{2}} by multiplying 1 by the reciprocal of \frac{\left(\sqrt{3}\right)^{2}}{2^{2}}.

$$\frac{4}{3}+\frac{2^{2}}{\left(\sqrt{3}\right)^{2}}-\left(\cos(45)\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{4}{3}+\frac{4}{\left(\sqrt{3}\right)^{2}}-\left(\cos(45)\right)^{2}$$

The square of \sqrt{3} is 3.

$$\frac{4}{3}+\frac{4}{3}-\left(\cos(45)\right)^{2}$$

Add \frac{4}{3} and \frac{4}{3} to get \frac{8}{3}.

$$\frac{8}{3}-\left(\cos(45)\right)^{2}$$

Get the value of \cos(45) from trigonometric values table.

$$\frac{8}{3}-\left(\frac{\sqrt{2}}{2}\right)^{2}$$

To raise \frac{\sqrt{2}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{8}{3}-\frac{\left(\sqrt{2}\right)^{2}}{2^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{8}{3}-\frac{2}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{8}{3}-\frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$\frac{8}{3}-\frac{1}{2}$$

Subtract \frac{1}{2} from \frac{8}{3} to get \frac{13}{6}.

$$\frac{13}{6}$$