Prove the trigonometric identity ?

#sin^2alpha (4cos^2alpha-1)^2+cos^2alpha(4sin^2alpha-1)^2=1#

1 Answer
Abhishek K.
Jun 3, 2018

#LHS=sin^2x(4cos^2x-1)^2+cos^2x(4sin^2x-1)^2#

#=sin^2x[16cos^4x-8cos^2x+1]+cos^2x[16sin^4x-8sin^2x+1]#

#=16cos^4x*sin^2x-8cos^2x*sin^2x+sin^2x+16sin^4x*cos^2x-8sin^2xcos^2x+cos^2x#

#=sin^2x+cos^2x+16cos^4x*sin^2x+16sin^4x*cos^2x-16sin^2x*cos^2x#

#=1+16sin^2xcos^2x[cos^2x+sin^2x-1]#

#=1+16sin^2xcos^2x[1-1]#

#=1+16sin^2xcos^2(x)xx0=1=RHS#

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