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

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

& 1+2+3+--+9=\dfrac{9\left( 9+1 \right)}{2} \\\ & \text{ }=\dfrac{9\times 10}{2} \\\ \end{aligned}$$ So we get $$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.