Question

Simplify the expression csc(-x)/1+tan^2x)

Answer 1

$\bf 1+tan^2(\theta)=sec^2(\theta)\qquad \qquad sin(-\theta )=-sin(\theta ) \\\\\\ cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)} \qquad csc(\theta)=\cfrac{1}{sin(\theta)} \qquad sec(\theta)=\cfrac{1}{cos(\theta)}\\\\ -------------------------------$

$\bf \cfrac{csc(-x)}{1+tan^2(x)}\implies \cfrac{\frac{1}{sin(-x)}}{sec^2(x)}\implies \cfrac{-\frac{1}{sin(x)}}{\frac{1}{cos^2(x)}}\implies -\cfrac{1}{sin(x)}\cdot \cfrac{cos^2(x)}{1} \\\\\\ -cos(x)\cdot \cfrac{cos(x)}{sin(x)}\implies -cos(x)cot(x)$

Answer 2

Assuming ya meant $\frac{csc(-x)}{1+tan^2(x)}$

to slimplify, we use a variation of the pythagorean identity and a decomposition into the sin and cos


for the pythaogreaon identity
$cos^2(x)+sin^2(x)=1$
divide both sides by $cos^2(x)$
$1+tan^2(x)=sec^2(x)$ since $\frac{sin(x)}{cos(x)}=tan(x)$
subsitute
$\frac{csc(x)}{sec^2(x)}$

recall that $csc(x)=\frac{1}{cos(x)}$
also that cos(x) is an even function and thus cos(-x)=cos(x)
therfore $csc(-x)=\frac{1}{cos(-x)}=\frac{1}{cos(x)}=csc(x)$
so we get

$\frac{csc(x)}{sec^2(x)}$
decompose them into $\frac{1}{cos(x)}$ and $\frac{1}{sin^2(x)}$ to get $\frac{\frac{1}{cos(x)}}{\frac{1}{sin^2(x)}}$
multiply by $\frac{sin^2(x)}{sin^2(x)}$ to get
$\frac{sin^2(x)}{cos(x)}$
we can furthur simlify to get
$(\frac{sin(x)}{cos(x)})(sin(x))=tan(x)sin(x)$
the expression simplifies to tan(x)sin(x)