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#