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Question: How do you simplify \(\tan \left( \pi +\theta \right)\)?...

How do you simplify tan⁡(π+θ)\tan \left( \pi +\theta \right)?

Explanation

Solution

We can solve the above question by addition of tan formula , we know that tan⁡(a+b)=tan⁡a+tan⁡b1−tan⁡atan⁡b\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b} so we can assume a is equal to π\pi and b is equal to θ\theta and replace these in the formula to solve the problem .

Complete step by step answer:
We have to simplify tan⁡(π+θ)\tan \left( \pi +\theta \right)
We know that tan⁡(a+b)=tan⁡a+tan⁡b1−tan⁡atan⁡b\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}
Replacing a with π\pi and b with θ\theta in the above formula we get
tan⁡(π+θ)=tan⁡π+tan⁡θ1−tan⁡πtan⁡θ\tan \left( \pi +\theta \right)=\dfrac{\tan \pi +\tan \theta }{1-\tan \pi \tan \theta }
We know that the value of tan⁡π\tan \pi is equal to 0 so putting in the place of tan⁡π\tan \pi we get
⇒tan⁡(π+θ)=0+tan⁡θ1−0×tan⁡θ\Rightarrow \tan \left( \pi +\theta \right)=\dfrac{0+\tan \theta }{1-0\times \tan \theta }
Further solving we get
⇒tan⁡(π+θ)=tan⁡θ\Rightarrow \tan \left( \pi +\theta \right)=\tan \theta

Note:
Another method we can apply to solve this problem we can write tan⁡(π+θ)\tan \left( \pi +\theta \right) as
sin⁡(π+θ)cos⁡(π+θ)\dfrac{\sin \left( \pi +\theta \right)}{\cos \left( \pi +\theta \right)} we can get the value of sin⁡(π+θ)\sin \left( \pi +\theta \right) and cos⁡(π+θ)\cos \left( \pi +\theta \right) by addition formula or geometric method , the value of sin⁡(π+θ)\sin \left( \pi +\theta \right) is equal to −sin⁡θ-\sin \theta and the value of −cos⁡θ-\cos \theta
So now we can write sin⁡(π+θ)cos⁡(π+θ)\dfrac{\sin \left( \pi +\theta \right)}{\cos \left( \pi +\theta \right)} = −sin⁡θ−cos⁡θ\dfrac{-\sin \theta }{-\cos \theta }
We can cancel out -1 in numerator and denominator
So sin⁡(π+θ)cos⁡(π+θ)\dfrac{\sin \left( \pi +\theta \right)}{\cos \left( \pi +\theta \right)} = tan⁡θ\tan \theta
So tan⁡(π+θ)=tan⁡θ\tan \left( \pi +\theta \right)=\tan \theta
We know that if a function f which has the property
f(x+c)=f(x)f\left( x+c \right)=f\left( x \right) where c is the smallest possible value then we say function f has a period of c
The graph of function f will repeat after a length of c units
We can compare function f to function tan x which has a property tan⁡(π+θ)=tan⁡θ\tan \left( \pi +\theta \right)=\tan \theta so we can say tan x has a period of π\pi and the graph of tan x will repeat itself after a length π\pi .
All the trigonometric function except tan x and cot x has period 2π2\pi ,tan x and cot x has period of π\pi