Question

Use the periodic and even–odd properties.

If $f\left(\theta \right)=\tan \theta$and $f\left(a\right)=2$, find the exact value of :

(a) $f(-a)$

(b) $f\left(a\right)+f(a+\pi )+f(a+2\pi )$

Short answer

(a)The value of $f(-a)$is$-2$.

(b) The value of $f\left(a\right)+f(a+\pi )+f(a+2\pi )$is$6$.

Step-by-step solution

Step 1. Given Information  

We have given that following function :-

$f\left(\theta \right)=\tan \theta$and $f\left(a\right)=2$.

We have to find the value of $f(-a)$and value of $f\left(a\right)+f(a+\pi )+f(a+2\pi )$.

To find value of $f(-a)$we will use even-odd properties and to find the value of$f\left(a\right)+f(a+\pi )+f(a+2\pi )$ we will use periodic properties.

Step 2. Part (a). To find value of f(-a)

We have given that :-

$f\left(\theta \right)=\tan \theta$and $f\left(a\right)=2$.

We know that :-

$\tan (-\theta )=-\tan \theta$.

Now put role="math" $\theta =-a$in $f\left(\theta \right)=\tan \theta$, then we have :-

$$\begin{gathered} f(-a)=\tan (-a) \\ \Rightarrow f(-a)=-\tan \left(a\right) \\ \Rightarrow f(-a)=-f\left(a\right) \end{gathered}$$

Put $f\left(a\right)=2$, then we have :-

$f(-a)=-2$.

This is the required value.

Step 3. Part (b). To find value of f(a)+f(a+π)+f(a+2π)

We have :-

$f\left(\theta \right)=\tan \theta$.

We know that tangent function is periodic of period $\pi$.

This gives us :-

$\tan (\theta +\pi k)=\tan \theta$.

Now :-

localid="1646925176572" $$\begin{gathered} f\left(a\right)+f(a+\pi )+f(a+2\pi )=\tan \left(a\right)+\tan (a+\pi )+\tan (a+2\pi ) \\ \Rightarrow f\left(a\right)+f(a+\pi )+f(a+2\pi )=\tan \left(a\right)+\tan \left(a\right)+\tan \left(a\right) \\ \Rightarrow f\left(a\right)+f(a+\pi )+f(a+2\pi )=3\tan \left(a\right) \\ \Rightarrow f\left(a\right)+f(a+\pi )+f(a+2\pi )=3f\left(a\right) \end{gathered}$$

Put $f\left(a\right)=2$, then we have :-

$$\begin{gathered} f\left(a\right)+f(a+\pi )+f(a+2\pi )=3\times 2 \\ \Rightarrow f\left(a\right)+f(a+\pi )+f(a+2\pi )=6 \end{gathered}$$

This is the required value.