Question

please solve the question urgent

Answer 1

L.H.s

=3[(sinx - cosx )^2]^2+4(sin^6x + cos^6x)+6(sinx +cosx )^2

=3(sin^2x +cos^2x -2cosxsinx)^2+4[(sin^2x )^3+ (cos^2x^3)]+6(sin^2x +cos^2x+2cosxsinx)

as ;(sin^2x )^3+ (cos^2x)^3=(sin^2x + cos^2x) (sin^4x + cos^4x – sin^2x cos^2x) so eq becomes

=3(sin^2x +cos^2x -2cosxsinx)^2+4(sin^2x + cos^2x) (sin^4x + cos^4x – sin^2x cos^2x)]+6(sin^2x +cos^2x+2cosxsinx)

=3(sin^2x +cos^2x -2cosxsinx)^2+4 [(sin^2x)^2 + (cos^2 x)^2 + 2 sin^2x cos^2x – 3 sin^2 x cos^2x]+6(sin^2x +cos^2x+2cosxsinx)

as we know sin^2x+cos^2x=1 so above eq becomes;

=3(1-2cosxsinx)^2+ 4 [(sin^2x + cos ^2x)^2 – 3 sin2x cos2x] + 6(1 - 2cosxsinx)

= 3 [1 + 4 sin^2x cos^2x – 4 sin x cos x)] + 4(1 - 3 sin2x cos2x)+ 6(1 - 2cosxsinx)

= 3 + 12 sin^2x cos^2x – 12 sin x cos x + 4 - 12 sin^2x cos^2x + 6 - 12cosxsinx

= 13          hence proved