Respuesta :
assuming ya meant [tex]\frac{csc(-x)}{1+tan^2(x)}[/tex]
to slimplify, we use a variation of the pythagorean identity and a decomposition into the sin and cos
for the pythaogreaon identity
[tex]cos^2(x)+sin^2(x)=1[/tex]
divide both sides by [tex]cos^2(x)[/tex]
[tex]1+tan^2(x)=sec^2(x)[/tex] since [tex] \frac{sin(x)}{cos(x)}=tan(x) [/tex]
subsitute
[tex]\frac{csc(x)}{sec^2(x)}[/tex]
recall that [tex]csc(x)=\frac{1}{cos(x)}[/tex]
also that cos(x) is an even function and thus cos(-x)=cos(x)
therfore [tex] csc(-x)=\frac{1}{cos(-x)}=\frac{1}{cos(x)}=csc(x)[/tex]
so we get
[tex]\frac{csc(x)}{sec^2(x)}[/tex]
decompose them into [tex]\frac{1}{cos(x)}[/tex] and [tex]\frac{1}{sin^2(x)}[/tex] to get [tex]\frac{\frac{1}{cos(x)}}{\frac{1}{sin^2(x)}}[/tex]
multiply by [tex]\frac{sin^2(x)}{sin^2(x)}[/tex] to get
[tex]\frac{sin^2(x)}{cos(x)}[/tex]
we can furthur simlify to get
[tex](\frac{sin(x)}{cos(x)})(sin(x))=tan(x)sin(x)[/tex]
the expression simplifies to tan(x)sin(x)
to slimplify, we use a variation of the pythagorean identity and a decomposition into the sin and cos
for the pythaogreaon identity
[tex]cos^2(x)+sin^2(x)=1[/tex]
divide both sides by [tex]cos^2(x)[/tex]
[tex]1+tan^2(x)=sec^2(x)[/tex] since [tex] \frac{sin(x)}{cos(x)}=tan(x) [/tex]
subsitute
[tex]\frac{csc(x)}{sec^2(x)}[/tex]
recall that [tex]csc(x)=\frac{1}{cos(x)}[/tex]
also that cos(x) is an even function and thus cos(-x)=cos(x)
therfore [tex] csc(-x)=\frac{1}{cos(-x)}=\frac{1}{cos(x)}=csc(x)[/tex]
so we get
[tex]\frac{csc(x)}{sec^2(x)}[/tex]
decompose them into [tex]\frac{1}{cos(x)}[/tex] and [tex]\frac{1}{sin^2(x)}[/tex] to get [tex]\frac{\frac{1}{cos(x)}}{\frac{1}{sin^2(x)}}[/tex]
multiply by [tex]\frac{sin^2(x)}{sin^2(x)}[/tex] to get
[tex]\frac{sin^2(x)}{cos(x)}[/tex]
we can furthur simlify to get
[tex](\frac{sin(x)}{cos(x)})(sin(x))=tan(x)sin(x)[/tex]
the expression simplifies to tan(x)sin(x)
[tex]\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)}\\\\
-------------------------------[/tex]
[tex]\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)[/tex]
[tex]\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)[/tex]
