Question

The expression ,{Tan(x-pi/2).cos(3pi/2+x)-sin^3(7pi/2-x)} / cos(x-pi/2).tan(3pi/2+x)Simplifies to,A) (1+cos^2x)B)sin^2xC)-(1+cos^2x)D)cos^2x

Answer 1

{tan(x-π/2).cos(3π/2+x)-sin³(7π/2-x)}/{cos(x-π/2)tan(3π/2+x)}
=[tan{-(π/2-x)}.cos{(π/2×3)+x}-sin³{(π/2×7)-x}]/[cos{-(π/2-x)}.tan{(π/2×3)+x}
=[-tan(π/2-x).sinx-(-cosx)³]/[cos(π/2-x).(-cotx)]
=(-cotx.sinx+cos³x)/[sinx.(-cotx)]
=[-(cosx/sinx).sinx+cos³x]/[sinx.(-cosx/sinx)]
={-cosx(1-cos²x)}/(-cosx)
=1-cos²x
=sin²x

Answer 2

Answer:

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