How do you prove $\tan \left( {{180}^{\circ }}-a \right)=-\tan a$ ?

Hint: We can prove $\tan \left( {{180}^{\circ }}-a \right)$ is equal to – tan a by the difference formula tan (x-y) = $\dfrac{\tan x-\tan y}{1+\tan x\tan y}$ where a and b are not equal to $\dfrac{\pi }{2}$ . We can assume x as 180 and y as a then apply the formula to solve the question.

Complete step by step answer:
We have to prove $\tan \left( {{180}^{\circ }}-a \right)=-\tan a$
We know that tan (x-y) = $\dfrac{\tan x-\tan y}{1+\tan x\tan y}$ so let’s apply the formula by taking x = ${{180}^{\circ }}$and y = a
$\Rightarrow \tan \left( {{180}^{\circ }}-a \right)=\dfrac{\tan {{180}^{\circ }}-\tan a}{1+\tan {{180}^{\circ }}\tan a}$
We know that the value of $\tan {{180}^{\circ }}$ is equal to 0.
Putting 0 in place of $\tan {{180}^{\circ }}$ we get
$\Rightarrow \tan \left( {{180}^{\circ }}-a \right)=\dfrac{0-\tan a}{1+0\times \tan a}$
Further solving we get
 $\Rightarrow \tan \left( {{180}^{\circ }}-a \right)=-\tan a$ where a is not equal to $\dfrac{n\pi }{2}$ ; n is an integer

Note:
We can prove $\tan \left( {{180}^{\circ }}-a \right)$ is equal to – tan a by another , we know that tan x is the ratio of sin x and cos x , so $\tan \left( {{180}^{\circ }}-a \right)$ is the ratio of $\sin \left( {{180}^{\circ }}-a \right)$ and $\cos \left( {{180}^{\circ }}-a \right)$ .
$\Rightarrow \tan \left( {{180}^{\circ }}-a \right)=\dfrac{\sin \left( {{180}^{\circ }}-a \right)}{\cos \left( {{180}^{\circ }}-a \right)}$ where a is not equal to $\dfrac{n\pi }{2}$
We know that the value of $\sin \left( {{180}^{\circ }}-a \right)$ is equal to sin a and the value of $\cos \left( {{180}^{\circ }}-a \right)$ is – cos a
so we can write $\tan \left( {{180}^{\circ }}-a \right)=\dfrac{\sin a}{-\cos a}$ where a is not equal to $\dfrac{n\pi }{2}$
Now we can see $\tan \left( {{180}^{\circ }}-a \right)$ is equal to – tan a , where a is not equal to $\dfrac{n\pi }{2}$
Always keep mentioning the angle where the trigonometric function does not exist, that is because the proof we are doing will not be valid at that angle. So don’t forget to mention that like we did in the above proof. We have not included $\dfrac{n\pi }{2}$ because at $\dfrac{n\pi }{2}$ the value of $\dfrac{n\pi }{2}$ tends to infinity.