If #sin x + sin^2 x = 1# then the value of #cos^2x + cos^4x + cot^4x - cot^2x# is?

Question

A) 1
B) 0
C) 2
D) None of these

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Answer 1 · 2017-08-18T08:00:21

Given

#sinx+sin^2x=1#

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

#=>sinx=cos^2x#

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

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

#=>cos^2x+cos^4x=1...... [1]#

Again

#sinx+sin^2x=1#

#=>sinx/sin^2x+sin^2x/sin^2x=1/sin^2x#

#=>cscx+1=csc^2x#

#=>cscx=csc^2x-1#

#=>cscx=cot^2x#

#=>csc^2x=cot^4x#

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

#=>cot^4x-cot^2x=1........ [2]#

Adding [1] and [2]

we get

#cos^2x+cos^4x+cot^4x-cot^2x=1+1=2#