Solve 3tanpi/1-tan^2frac{pi{8}}=

$$\frac { 3 \tan \pi } { 1 - \tan ^ { 2 } \frac { \pi } { 8 } } =$$

Evaluate

$$0$$

Solution Steps

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

$$\frac{3\times 0}{1-\left(\tan(\frac{\pi }{8})\right)^{2}}$$

Multiply 3 and 0 to get 0.

$$\frac{0}{1-\left(\tan(\frac{\pi }{8})\right)^{2}}$$

Get the value of \tan(\frac{\pi }{8}) from trigonometric values table.

$$\frac{0}{1-\left(\sqrt{2}-1\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{2}-1\right)^{2}.

$$\frac{0}{1-\left(\left(\sqrt{2}\right)^{2}-2\sqrt{2}+1\right)}$$

The square of \sqrt{2} is 2.

$$\frac{0}{1-\left(2-2\sqrt{2}+1\right)}$$

Add 2 and 1 to get 3.

$$\frac{0}{1-\left(3-2\sqrt{2}\right)}$$

To find the opposite of 3-2\sqrt{2}, find the opposite of each term.

$$\frac{0}{1-3+2\sqrt{2}}$$

Subtract 3 from 1 to get -2.

$$\frac{0}{-2+2\sqrt{2}}$$

Zero divided by any non-zero term gives zero.

$$0$$

Factor

$$0$$