Question

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

Answer 1

assuming ya meant
`(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
`(sin(x))/(cos(x))=tan(x)`
subsitute

`(csc(x))/(sec^2(x))`

recall that
`csc(x)=(1)/(cos(x))`
also that cos(x) is an even function and thus cos(-x)=cos(x)
therfore
`csc(-x)=(1)/(cos(-x))=(1)/(cos(x))=csc(x)`
so we get


`(csc(x))/(sec^2(x))`
decompose them into
`(1)/(cos(x))` and
`(1)/(sin^2(x))` to get
`((1)/(cos(x)))/((1)/(sin^2(x)))`
multiply by
`(sin^2(x))/(sin^2(x))` to get

`(sin^2(x))/(cos(x))`
we can furthur simlify to get

`((sin(x))/(cos(x)))(sin(x))=tan(x)sin(x)`
the expression simplifies to tan(x)sin(x)

Answer 2

`\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{(1)/(sin(-x))}{sec^2(x)}\implies \cfrac{-(1)/(sin(x))}{(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)`