How do you simplify $\tan \left( \pi +\theta \right)$?
Answer
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Hint: We can solve the above question by addition of tan formula , we know that $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$ so we can assume a is equal to $\pi $ and b is equal to $\theta $ and replace these in the formula to solve the problem .
Complete step by step answer:
We have to simplify $\tan \left( \pi +\theta \right)$
We know that $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$
Replacing a with $\pi $ and b with $\theta $ in the above formula we get
$\tan \left( \pi +\theta \right)=\dfrac{\tan \pi +\tan \theta }{1-\tan \pi \tan \theta }$
We know that the value of $\tan \pi $ is equal to 0 so putting in the place of $\tan \pi $ we get
$\Rightarrow \tan \left( \pi +\theta \right)=\dfrac{0+\tan \theta }{1-0\times \tan \theta }$
Further solving we get
$\Rightarrow \tan \left( \pi +\theta \right)=\tan \theta $
Note:
Another method we can apply to solve this problem we can write $\tan \left( \pi +\theta \right)$ as
$\dfrac{\sin \left( \pi +\theta \right)}{\cos \left( \pi +\theta \right)}$ we can get the value of $\sin \left( \pi +\theta \right)$ and $\cos \left( \pi +\theta \right)$ by addition formula or geometric method , the value of $\sin \left( \pi +\theta \right)$ is equal to $-\sin \theta $ and the value of $-\cos \theta $
So now we can write $\dfrac{\sin \left( \pi +\theta \right)}{\cos \left( \pi +\theta \right)}$ = $\dfrac{-\sin \theta }{-\cos \theta }$
We can cancel out -1 in numerator and denominator
So $\dfrac{\sin \left( \pi +\theta \right)}{\cos \left( \pi +\theta \right)}$ = $\tan \theta $
So $\tan \left( \pi +\theta \right)=\tan \theta $
We know that if a function f which has the property
$f\left( x+c \right)=f\left( x \right)$ 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 \left( \pi +\theta \right)=\tan \theta $ so we can say tan x has a period of $\pi $ and the graph of tan x will repeat itself after a length $\pi $ .
All the trigonometric function except tan x and cot x has period $2\pi $ ,tan x and cot x has period of $\pi $
Complete step by step answer:
We have to simplify $\tan \left( \pi +\theta \right)$
We know that $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$
Replacing a with $\pi $ and b with $\theta $ in the above formula we get
$\tan \left( \pi +\theta \right)=\dfrac{\tan \pi +\tan \theta }{1-\tan \pi \tan \theta }$
We know that the value of $\tan \pi $ is equal to 0 so putting in the place of $\tan \pi $ we get
$\Rightarrow \tan \left( \pi +\theta \right)=\dfrac{0+\tan \theta }{1-0\times \tan \theta }$
Further solving we get
$\Rightarrow \tan \left( \pi +\theta \right)=\tan \theta $
Note:
Another method we can apply to solve this problem we can write $\tan \left( \pi +\theta \right)$ as
$\dfrac{\sin \left( \pi +\theta \right)}{\cos \left( \pi +\theta \right)}$ we can get the value of $\sin \left( \pi +\theta \right)$ and $\cos \left( \pi +\theta \right)$ by addition formula or geometric method , the value of $\sin \left( \pi +\theta \right)$ is equal to $-\sin \theta $ and the value of $-\cos \theta $
So now we can write $\dfrac{\sin \left( \pi +\theta \right)}{\cos \left( \pi +\theta \right)}$ = $\dfrac{-\sin \theta }{-\cos \theta }$
We can cancel out -1 in numerator and denominator
So $\dfrac{\sin \left( \pi +\theta \right)}{\cos \left( \pi +\theta \right)}$ = $\tan \theta $
So $\tan \left( \pi +\theta \right)=\tan \theta $
We know that if a function f which has the property
$f\left( x+c \right)=f\left( x \right)$ 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 \left( \pi +\theta \right)=\tan \theta $ so we can say tan x has a period of $\pi $ and the graph of tan x will repeat itself after a length $\pi $ .
All the trigonometric function except tan x and cot x has period $2\pi $ ,tan x and cot x has period of $\pi $
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