Question #5d572

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Answer 1 · 2016-12-26T07:23:29

Using the definitions of $csc$ and $cot$ , along with the identities

we have

$csc(2x)+cot(2x) = 1/sin(2x)+cos(2x)/sin(2x)$

$=(1+cos(2x))/sin(2x)$

$=(1+(2cos^2(x)-1))/(2sin(x)cos(x))$

$=(2cos^2(x))/(2sin(x)cos(x))$

$=cos(x)/sin(x)$

$=cot(x)$

Answer 2 · 2016-12-26T07:40:12

See proof below

We use

$cscx=1/sinx$

$sin2x=2sinxcosx$

$cos2x=2cos^2x-1$

$cotx=cosx/sinx$

So,

$csc2x+cot2x=1/(sin2x)+(cos2x)/(sin2x)$

$=(1+cos2x)/(sin2x)$

$=(1+cos^2x-1)/(2sinxcosx)$

$=(2cos^2x)/(2sinxcosx)$

$=cosx/sinx$

$=cotx$

Q.E.D