Solve cos 60 deg = (1 - tan^2 (30 deg))/(1 + tan^2 (30 deg)) tan 30 ° | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\cos 60 ^ { \circ } = \frac { 1 - \tan ^ { 2 } 30 ^ { \circ } } { 1 + \tan ^ { 2 } 30 ^ { \circ } }$$

Verify

$\text{true}$

Solution Steps

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

$$\frac{1}{2}=\frac{1-\left(\tan(30)\right)^{2}}{1+\left(\tan(30)\right)^{2}}$$

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

$$\frac{1}{2}=\frac{1-\left(\frac{\sqrt{3}}{3}\right)^{2}}{1+\left(\tan(30)\right)^{2}}$$

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

$$\frac{1}{2}=\frac{1-\frac{\left(\sqrt{3}\right)^{2}}{3^{2}}}{1+\left(\tan(30)\right)^{2}}$$

The square of $\sqrt{3}$ is $3$.

$$\frac{1}{2}=\frac{1-\frac{3}{3^{2}}}{1+\left(\tan(30)\right)^{2}}$$

Calculate $3$ to the power of $2$ and get $9$.

$$\frac{1}{2}=\frac{1-\frac{3}{9}}{1+\left(\tan(30)\right)^{2}}$$

Reduce the fraction $\frac{3}{9}$ to lowest terms by extracting and canceling out $3$.

$$\frac{1}{2}=\frac{1-\frac{1}{3}}{1+\left(\tan(30)\right)^{2}}$$

Subtract $\frac{1}{3}$ from $1$ to get $\frac{2}{3}$.

$$\frac{1}{2}=\frac{\frac{2}{3}}{1+\left(\tan(30)\right)^{2}}$$

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

$$\frac{1}{2}=\frac{\frac{2}{3}}{1+\left(\frac{\sqrt{3}}{3}\right)^{2}}$$

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

$$\frac{1}{2}=\frac{\frac{2}{3}}{1+\frac{\left(\sqrt{3}\right)^{2}}{3^{2}}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply $1$ times $\frac{3^{2}}{3^{2}}$.

$$\frac{1}{2}=\frac{\frac{2}{3}}{\frac{3^{2}}{3^{2}}+\frac{\left(\sqrt{3}\right)^{2}}{3^{2}}}$$

Since $\frac{3^{2}}{3^{2}}$ and $\frac{\left(\sqrt{3}\right)^{2}}{3^{2}}$ have the same denominator, add them by adding their numerators.

$$\frac{1}{2}=\frac{\frac{2}{3}}{\frac{3^{2}+\left(\sqrt{3}\right)^{2}}{3^{2}}}$$

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

$$\frac{1}{2}=\frac{2\times 3^{2}}{3\left(3^{2}+\left(\sqrt{3}\right)^{2}\right)}$$

Cancel out $3$ in both numerator and denominator.

$$\frac{1}{2}=\frac{2\times 3}{\left(\sqrt{3}\right)^{2}+3^{2}}$$

Multiply $2$ and $3$ to get $6$.

$$\frac{1}{2}=\frac{6}{\left(\sqrt{3}\right)^{2}+3^{2}}$$

The square of $\sqrt{3}$ is $3$.

$$\frac{1}{2}=\frac{6}{3+3^{2}}$$

Calculate $3$ to the power of $2$ and get $9$.

$$\frac{1}{2}=\frac{6}{3+9}$$

Add $3$ and $9$ to get $12$.

$$\frac{1}{2}=\frac{6}{12}$$

Reduce the fraction $\frac{6}{12}$ to lowest terms by extracting and canceling out $6$.

$$\frac{1}{2}=\frac{1}{2}$$

Compare $\frac{1}{2}$ and $\frac{1}{2}$.

$$\text{true}$$