Check my work: Evaluating $\tan\frac{7\pi}{8}$ using a half-angle formula

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I am doing a trig problem involving half-angle identities, and I am not sure if my solution is correct. Can someone please check my work?

The question:

Find the exact value of $\tan\frac{7\pi}{8}$ by using a half-angle identity.

My work:

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2 Answers

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Observe that $$\tan \frac{7\pi}{8}=-\tan\left( \pi -\frac{7\pi}{8} \right)$$

So $$\tan \frac{7\pi}{8}=-\tan\frac{\pi}{8} $$

Now use $$\tan 2\theta =\frac{2\tan \theta }{1-\tan^2 \theta}$$

Let $$\theta = \frac{\pi}{8}$$

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

$$\tan \frac{\pi}{4} =1=\frac{2\tan \frac{\pi}{8} }{1-\tan^2 \frac{\pi}{8}}$$

Put $x=\tan \frac{\pi}{8}$

Then $$1=\frac{2x}{1-x^2}$$

$$1-x^2=2x$$

$$x^2+2x-1=0$$

$$x=-1-\sqrt{2}$$

$$x=\sqrt{2}-1$$

Since $$\tan \frac{\pi}{8} >0$$

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

Therefore $$\tan \frac{7\pi}{8}=-\tan\frac{\pi}{8}= -(\sqrt{2}-1) $$

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It's easier to use $$\tan\dfrac{7\pi}8=\dfrac{1-\cos\dfrac{7\pi}4}{\sin\dfrac{7\pi}4}$$

Now $\sin\dfrac{7\pi}4=\sin\left(2\pi-\dfrac\pi4\right)=-\sin\dfrac\pi4=-\dfrac1{\sqrt2}$

and $\cos\dfrac{7\pi}4=\cos\left(2\pi-\dfrac\pi4\right)=\cos\dfrac\pi4=+\dfrac1{\sqrt2}$

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