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

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

Explanation

Solution

We can prove tan⁡(180∘−a)\tan \left( {{180}^{\circ }}-a \right) is equal to – tan a by the difference formula tan (x-y) = tan⁡x−tan⁡y1+tan⁡xtan⁡y\dfrac{\tan x-\tan y}{1+\tan x\tan y} where a and b are not equal to π2\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⁡(180∘−a)=−tan⁡a\tan \left( {{180}^{\circ }}-a \right)=-\tan a
We know that tan (x-y) = tan⁡x−tan⁡y1+tan⁡xtan⁡y\dfrac{\tan x-\tan y}{1+\tan x\tan y} so let’s apply the formula by taking x = 180∘{{180}^{\circ }}and y = a
⇒tan⁡(180∘−a)=tan⁡180∘−tan⁡a1+tan⁡180∘tan⁡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∘\tan {{180}^{\circ }} is equal to 0.
Putting 0 in place of tan⁡180∘\tan {{180}^{\circ }} we get
⇒tan⁡(180∘−a)=0−tan⁡a1+0×tan⁡a\Rightarrow \tan \left( {{180}^{\circ }}-a \right)=\dfrac{0-\tan a}{1+0\times \tan a}
Further solving we get
⇒tan⁡(180∘−a)=−tan⁡a\Rightarrow \tan \left( {{180}^{\circ }}-a \right)=-\tan a where a is not equal to nπ2\dfrac{n\pi }{2} ; n is an integer

Note:
We can prove tan⁡(180∘−a)\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⁡(180∘−a)\tan \left( {{180}^{\circ }}-a \right) is the ratio of sin⁡(180∘−a)\sin \left( {{180}^{\circ }}-a \right) and cos⁡(180∘−a)\cos \left( {{180}^{\circ }}-a \right) .
⇒tan⁡(180∘−a)=sin⁡(180∘−a)cos⁡(180∘−a)\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 nπ2\dfrac{n\pi }{2}
We know that the value of sin⁡(180∘−a)\sin \left( {{180}^{\circ }}-a \right) is equal to sin a and the value of cos⁡(180∘−a)\cos \left( {{180}^{\circ }}-a \right) is – cos a
so we can write tan⁡(180∘−a)=sin⁡a−cos⁡a\tan \left( {{180}^{\circ }}-a \right)=\dfrac{\sin a}{-\cos a} where a is not equal to nπ2\dfrac{n\pi }{2}
Now we can see tan⁡(180∘−a)\tan \left( {{180}^{\circ }}-a \right) is equal to – tan a , where a is not equal to nπ2\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 nπ2\dfrac{n\pi }{2} because at nπ2\dfrac{n\pi }{2} the value of nπ2\dfrac{n\pi }{2} tends to infinity.