Question

How do you simplify \[\tan \left( x+\pi \right)\]?

Answer 1

Hint: Assume the given expression as ‘E’. Consider x = A and \[\pi =B\] and use the formula for tangent of sum of two angles given as: - \[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\]. Substitute back the assumed values of A and B and use the value of \[\tan \pi =0\] to simplify the expression. Simplify both the numerator and denominator using the above value and get the answer.

Complete step by step solution:
Here, we have been provided with the trigonometric expression \[\tan \left( x+\pi \right)\] and we are asked to simplify it. Let us assume the given expression as ‘E’, so we have,
\[\Rightarrow E=\tan \left( x+\pi \right)\]
Here, we can see that we have a sum of two angles x and \[\pi \] whose value of the tangent function we have to determine. Now, assuming x = A and \[\pi =B\], we have the expression,
\[\Rightarrow E=\tan \left( A+B \right)\]
Using the identity: - \[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\], we get,
\[\Rightarrow E=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\]
Substituting back the values of A and B, we get,
\[\Rightarrow E=\dfrac{\tan x+\tan \pi }{1-\tan x.\tan \pi }\]
We know that the value of the tangent of integral multiple of \[\pi \] is 0. This is because tangent function is the ratio of sine function to the cosine function and the value of sine of integral multiple of \[\pi \] is 0. So, mathematically, we have,
\[\Rightarrow \sin \left( n\pi \right)=0,n\in \] Integers
\[\Rightarrow \tan \left( n\pi \right)=\dfrac{\sin \left( n\pi \right)}{\cos \left( n\pi \right)},n\in \] Integers
Here, for n = 1 we have, \[\tan \pi =0\]. Substituting this value in the above obtained expression ‘E’, we get,
\[\Rightarrow E=\dfrac{\tan x+0}{1-\tan x.0}\]
\[\begin{align}
  & \Rightarrow E=\dfrac{\tan x}{1} \\
 & \Rightarrow E=\tan x \\
\end{align}\]
Hence, the simplified value of the given expression is \[\tan x\], which is our answer.

Note:
One may note that we can also solve this question without using the formula that we have applied in the solution. Since, we know that the period of the tangent function is \[\pi \], that means the value of the tangent function starts repeating itself after an interval of \[\pi \]. Mathematically, we can say that \[\tan \left( n\pi +\theta \right)=\tan \theta \], where \[n\in \] integers. So, we can directly write \[\tan \left( \pi +x \right)=\tan x\]. You can check the period of the tangent function by drawing its graph.