Question

Trig I need to simplify and write the trig expressions in terms of sin and cos

Answer 1

we have the expression

`(2+tan^2x)/(sec^2x)-1`

Remember that

`tan^2x+1=sec^2x`

substitute in the given expression

`\begin{gathered} (2+tan^(2)x)/(sec^(2)x)-1 \\ \\ (1+(1+tan^2x))/(sec^2x)-1 \\ \\ (1+(sec^2x))/(sec^2x)-1 \\ \\ (1+sec^2x-sec^2x)/(sec^2x) \\ \\ (1)/(sec^2x) \\ \\ cos^2x \end{gathered}`

therefore

`g(x)=cos^2x`