Question

`sin^6 theta + cos^6 theta =1 -3 sin^2 theta cos^2 theta`

Answer 1

LHS = `sin^6 theta + cos^6 theta`

=` (sin^2 theta )^3 + (cos^2 theta )^3`

Put sin² θ = a and cos² θ = b

∴ L.H.S. = a³ + b³

= (a + b)³ − 2ab (a + b)

=`(sin ^2 theta + cos^2 theta )-3 sin^2thetacos^2 theta(sin^2 theta cos^2 theta)`

=`(1)^3 - 3 sin^2 theta cos^2 theta`  [∴ sin² θ + cos² θ = 1]

=`1-3sin^2 theta cos^2 theta`

= RHS

Hence, LHS = RHS