How do i prove that $csc 2x+cot 2x=cot x$ ?

Question

I tried doing it:

$1/(sin2x)+1/(tan2x)$

$=1/(2sinx cost)+1/(2tan x/(1-tan^2 x)$

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

Can someone help please?

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Answer 1 · 2017-11-21T16:05:31

See the proof below

We need

$cos2x=1-2sin^2x$

$tan2x=(sin2x)/(cos2x)$

$cotx=cosx/sinx$

$cscx=1/sinx$

Therefore,

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

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

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

$=(2(1-sin^2x))/(2sinxcosx)$

$=cos^2x/(sinxcosx)$

$=cosx/sinx$

$=cotx$

$=RHS$

$QED$