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Answer 1 · 2018-01-12T05:43:48

#cot^4x-cot^2x=2-sqrt5#

.

#sin^4x+sin^2x=1#

#sin^4x+sin^2x-1=0#

#sin^2x=(-1+-sqrt(1-4(1)(-1)))/2=(-1+-sqrt5)/2#

We have to ignore the negative answer because #sin^2x# can only be positive.

#sin^2x=(-1+sqrt5)/2#

#cos^2x=1-sin^2x=1-(-1+sqrt5)/2=(2+1-sqrt5)/2=(3-sqrt5)/2#

#cot^2x=cos^2x/sin^2x=(3-sqrt5)/(-1+sqrt5)=((3-sqrt5)(sqrt5+1))/((sqrt5-1)(sqrt5+1))#

#cot^2x=(sqrt5-1)/2#

#cot^4x=((sqrt5-1)/2)^2=(6-2sqrt5)/4=(3-sqrt5)/2#

#cot^4x-cot^2x=(3-sqrt5)/2-(sqrt5-1)/2=(3-sqrt5-sqrt5+1)/2=(4-2sqrt5)/2=2-sqrt5#

Answer 2 · 2018-01-23T11:30:30

#sin^4 x+sin^2 x=1 #

#=>sin^4 x=1-sin^2x #

#=>sin^4 x=cos^2x #

#=>sin^4x/sin^4x=cos^2x/sin^4x #

#=>1=cot^2x xx csc^2x #

#=>1=cot^2x xx (1+cot^2x) #

#color(red)(=>cot^4x+cot^2x=1)#

#=>cot^4x+cot^2x-1=0#

#Cot^2x=(-1+sqrt(1+4))/2=(sqrt 5-1)/2#

Other root is negative and #cotx# becomes imaginary. So the negative value of #cot^2x# is neglected.

Now #cot^4x-cot^2x#

#=((sqrt5-1)/2)^2-(sqrt 5-1)/2#

#=1/4(6-2sqrt5)-1/4(2sqrt 5-2)#

#=1/4(6-2sqrt5-2sqrt 5+2)#

#=1/4(8-4sqrt5)#

#=2-sqrt5#