Question

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

Answer 1

we have the expression

$\frac{2+tan^2x}{sec^2x}-1$

Remember that

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

substitute in the given expression

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

therefore

$g(x)=cos^2x$