Question

6
State whether the function f(x)=cosx\1+sin^2x is even or odd


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Answer 1

$f(x)=\frac{cos(x)}{1+sin^2(x)}$  is an odd function.

Recall that the definition of an even function is

$f(x)=f(-x)$

and the definition of an odd function is

$f(x)=-f(x)$

Let's check either of these properties for our function

$f(x)=\frac{cos(x)}{1+sin^2(x)}$

taking into account that cos(x) is an even function because

$cos(x)=cos(-x)$

and sin(x) is an odd function because

$sin(-x)=-sin(x)$

Thus,

$f(-x)=\frac{cos(-x)}{1+sin^2(-x)}$

          $=\frac{cosx}{1+(-sin x)^2}$

          $=\frac{cosx}{1+sin^2x}$

Therefore, $f(x)=\frac{cos(x)}{1+sin^2(x)}$  is an odd function.

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Answer 2

Step-by-step explanation:

State whether the function f(x) = (cos x)/(1 + sin x) is even or odd function.