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.
$\endgroup$ 43 Answers
$\begingroup$Since $\sin(\pi)=0, \cos(\pi) = -1$ you have $$\tan(\pi - x) = \frac{\sin(\pi - x)}{\cos(\pi - x)} = \frac{\sin(\pi)\cos(x) - \sin(x)\cos(\pi)}{\cos(\pi)\cos(x) + \sin(\pi)\sin(x)} = \frac{+\sin(x)}{-\cos(x)} = -\frac{\sin(x)}{\cos(x)} = -\tan(x)$$
$\endgroup$ 1 $\begingroup$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)
$\endgroup$ 3 $\begingroup$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 $x$
