# Deduce $\tan(\pi - x) = -\tan x$

I have been trying to solve this problem:

"Use sum and difference identities and the sine and cosine functions to deduce that:"

$$\tan(\pi - x) = -\tan(x)$$

I can see that:

$$\tan(\pi - x) = \frac{\sin(\pi - x)}{\cos(\pi - x)} = \frac{\sin(\pi)\cos(x) - \sin(x)\cos(\pi)}{\cos(\pi)\cos(x) + \sin(\pi)\sin(x)}$$

But I have no idea where to go from here. I can't solve this algebra after a long time and really need to ask where to from here.

• Hint: $\sin(\pi)=0$ and $\cos(\pi)=-1$. Commented Nov 3, 2015 at 13:32
• That's done it! Commented Nov 3, 2015 at 13:37
• I beg you pardon, but $\sin(\pi-x)=\sin x$ and $\cos(\pi-x)=-\cos x$ for all $x$ are basic facts about sine and cosine. Commented Nov 3, 2015 at 13:38
• True, @Bernard. However, the OP was instructed to do it using the sine and cosine sum and difference identities. I guess they could do it by showing your identities as an intermediate step, though. Commented Nov 3, 2015 at 13:41

Since $\sin(\pi)=0, \cos(\pi) = -1$ you have $$\tan(\pi - x) = \frac{\sin(\pi - x)}{\cos(\pi - x)} = \frac{\sin(\pi)\cos(x) - \sin(x)\cos(\pi)}{\cos(\pi)\cos(x) + \sin(\pi)\sin(x)} = \frac{+\sin(x)}{-\cos(x)} = -\frac{\sin(x)}{\cos(x)} = -\tan(x)$$

• Thank you, I can now see that very clearly. Commented Nov 3, 2015 at 13:38

Hint

$\tan(A-B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}$. Use $A = \pi$ and $B = x$

So $\frac{\tan \pi-\tan x}{1 + \tan π\tan x} =-\tan x$ . (Since $\tan \pi = 0$).

So $\tan (\pi - x) = -\tan x$ (proved)

• You have to learn to format your answers. Otherwise they will get downvoted. You have been there on this site for a while now Commented Nov 3, 2015 at 13:46
• ya sure will try to improve. Commented Nov 3, 2015 at 13:58
• Also note that you had the formula wrong. The denominator has a + sign Commented Nov 3, 2015 at 14:18

Hint:
You can use this property, $$\cos(\pi-x)=-\cos x$$ Or by using this identity, $$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$ Substitute $A$ for $\pi$ and B for $x$