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

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2 votes
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)
User Thadeuszlay
by
9.3k points
2 votes

\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)
User James Moger
by
8.6k points

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