91Ó°ÊÓ

We have given that following function :-

$f\left(\mathrm{θ}\right)=\tan \mathrm{θ}$and $f\left(a\right)=2$.

We have to find the value of $f(-a)$and value of $f\left(a\right)+f(a+\mathrm{Ï€})+f(a+2\mathrm{Ï€})$.

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

We have given that :-

$f\left(\mathrm{θ}\right)=\tan \mathrm{θ}$and $f\left(a\right)=2$.

We know that :-

$\tan (-\mathrm{θ})=-\tan \mathrm{θ}$.

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

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

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

$f(-a)=-2$.

This is the required value.

We have :-

$f\left(\mathrm{θ}\right)=\tan \mathrm{θ}$.

We know that tangent function is periodic of period $\mathrm{Ï€}$.

This gives us :-

$\tan (\mathrm{θ}+\mathrm{π}k)=\tan \mathrm{θ}$.

Now :-

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

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

$$\begin{gathered} f\left(a\right)+f(a+\mathrm{π})+f(a+2\mathrm{π})=3×2 \\ ⇒f\left(a\right)+f(a+\mathrm{π})+f(a+2\mathrm{π})=6 \end{gathered}$$

This is the required value.