Q 115

Question

Use the periodic and even–odd properties.  

If f(θ)=tanθ and f(a)=2, find the exact value of :

(a) f(-a)

(b) f(a)+f(a+π)+f(a+2π)

Step-by-Step Solution

Verified
Answer

(a) The value of f(-a) is -2.

(b) The value of f(a)+f(a+π)+f(a+2π) is 6.

1Step 1. Given Information

We have given that following function :-  

f(θ)=tanθ and f(a)=2.

We have to find the value of f(-a) and value of f(a)+f(a+π)+f(a+2π).

To find value of f(-a) we will use even-odd properties and to find the value of f(a)+f(a+π)+f(a+2π) we will use periodic properties.

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

We have given that :- 

f(θ)=tanθ and f(a)=2.

We know that :-

tan(-θ)=-tanθ.

Now put θ=-a in f(θ)=tanθ, then we have :-

f(-a)=tan(-a)f(-a)=-tan(a)f(-a)=-f(a)

Put f(a)=2, then we have :-

f(-a)=-2.

This is the required value.

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

We have :-

f(θ)=tanθ.

We know that tangent function is periodic of period π.

This gives us :-

tan(θ+πk)=tanθ.

Now :-

f(a)+f(a+π)+f(a+2π)=tan(a)+tan(a+π)+tan(a+2π)f(a)+f(a+π)+f(a+2π)=tan(a)+tan(a)+tan(a)f(a)+f(a+π)+f(a+2π)=3tan(a)f(a)+f(a+π)+f(a+2π)=3f(a)

Put f(a)=2, then we have :-

f(a)+f(a+π)+f(a+2π)=3×2f(a)+f(a+π)+f(a+2π)=6

This is the required value.