Show that #(sinx)/(1 - cos2x) + (cosx)/(1 + cos2x) = (sinx+cosx)/(sin2x)#?

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Answer 1 · 2017-06-03T05:24:49

Try using these identities:

#sin^2x = (1 - cos2x)/2#

#cos^2x = (1 + cos2x)/2#

Then, we'd get:

#sinx/(1-cos2x) + cosx/(1+cos2x)#

#= sinx/(2sin^2x) + cosx/(2cos^2x)#

#= 1/(2sinx) + 1/(2cosx)#

We could try getting common denominators.

#= (cosx)/(2sinxcosx) + (sinx)/(2sinxcosx)#

Looks like that worked, because:

#sin2x = 2sinxcosx#.

Therefore:

#=> (sinx + cosx)/(2sinxcosx)#

#= color(blue)((sinx + cosx)/(sin2x))#

Maybe try verifying it from the right-hand side?

Answer 2 · 2017-06-03T05:26:14

See the proof below

We need

#cos2x=2cos^2x-1#

#cos2x=1-2sin^2x#

#2sinxcosx=sin2x#

Therefore,

#LHS=sinx/(1-cos2x)+cosx/(1+cos2x)#

#=sinx/(1-1+2sin^2x)+cosx/(1+2cos^2x-1)#

#=sinx/(2sin^2x)+cosx/(2cos^2x)#

#=1/(2sinx)+1/(2cosx)#

#=(cosx+sinx)/(2sinxcosx)#

#=(sinx+cosx)/(sin2x)#

#=RHS#

#QED#