Question

What is the period of the given function : $\tan \left( {{x}^{\circ }}+2{{x}^{\circ }}+--+9{{x}^{\circ }} \right)$
(a) $4$
(b) $8$
(c) $12$
(d) $2$

Answer 1

Hint: We are asked to find period of $\tan \left( {{x}^{\circ }}+2{{x}^{\circ }}+--+9{{x}^{\circ }} \right)$ , to find the period here we first have to simplify the angle inside $\tan $ . It is in the form of a series with n consecutive terms. So, we can use the formula $\dfrac{n\left( n+1 \right)}{2}$. Then, we will use the fact that the period for \[\tan \left( {{x}^{\circ }} \right)\] is \[{{180}^{\circ }}\] .

Complete step by step answer:
We are given the function as
$\tan \left( {{x}^{\circ }}+2{{x}^{\circ }}+--+9{{x}^{\circ }} \right)$
The terms inside the tangent function are in the form of consecutive terms.
We know sum of $n$ consecutive term is given by the formula $\dfrac{n\left( n+1 \right)}{2}$
Our term is ${{x}^{\circ }}+2{{x}^{\circ }}+--+9{{x}^{\circ }}$
Taking ${{x}^{\circ }}$ common , we get
 $\text{ }{{x}^{\circ }}\left( 1+2+3+--+9 \right)$
So it contains a sum of $9$ consecutive term. So applying the formula, we have
 \[1+2+3+--+9=\dfrac{n\left( n+1 \right)}{2}\] with \[n\] as \[9\]
So putting \[n\] as \[9\] and simplifying we get
 \[\begin{align}
  & 1+2+3+--+9=\dfrac{9\left( 9+1 \right)}{2} \\
 & \text{ }=\dfrac{9\times 10}{2} \\
\end{align}\]
So we get
 \[1+2+3+--+9=45\]
Hence we can substitute it as
 \[{{x}^{\circ }}+2{{x}^{\circ }}+3{{x}^{\circ }}+--+9{{x}^{\circ }}=45{{x}^{\circ }}\]
So our initial trigonometric ratio become
 \[\tan \left( {{x}^{\circ }}+2{{x}^{\circ }}+3{{x}^{\circ }}+--+9{{x}^{\circ }} \right)=\tan \left( 45{{x}^{\circ }} \right)\]
We know for \[\tan \left( {{x}^{\circ }} \right)\] period is \[{{180}^{\circ }}\]
And for \[\tan \left( \alpha {{x}^{\circ }} \right)\] period is given by \[\dfrac{{{180}^{\circ }}}{\alpha }\] .
For our function, we have
 \[\tan \left( {{x}^{\circ }}+2{{x}^{\circ }}+3{{x}^{\circ }}+--+9{{x}^{\circ }} \right)=\tan \left( 45{{x}^{\circ }} \right)\] So \[\alpha \] is \[{{45}^{\circ }}\]
So period of $\tan \left( {{x}^{\circ }}+2{{x}^{\circ }}+--+9{{x}^{\circ }} \right)$ will be \[\dfrac{{{180}^{\circ }}}{{{45}^{\circ }}}\]
Simplifying we get
Period of $\tan \left( {{x}^{\circ }}+2{{x}^{\circ }}+--+9{{x}^{\circ }} \right)=\dfrac{{{180}^{\circ }}}{{{45}^{\circ }}}=4$

So, the correct answer is “Option A”.

Note: Using sum of $n$ consecutive term formula as $\dfrac{n\left( n+1 \right)}{2}$ will reduce the effort and improve the efficiency as if we count it without formula it take a lot of time and for higher numbers it is impossible to calculate by normal summation as our term more from $1$ and goes upto $9$ so there are total of $9$ terms are so we use $9$ and find sum.