# 2(sin6θ + cos6θ) − 3(sin4θ + cos4θ) is equal toA. 0B. 1C. −1D. None of these

To find: 2(sin6 θ + cos6 θ) – 3(sin4 θ + cos4 θ)

First, we consider

sin6 θ + cos6 θ = (sin2 θ)3 + (cos2 θ)3

Now, as (a + b)3 = a3 + b3 + 3a2b + 3ab2

a3 + b3 = (a + b)3 – 3a2b – 3ab2

sin6 θ + cos6 θ

= (sin2 θ)3 + (cos2 θ)3

= (sin2 θ + cos2 θ)3 – 3 (sin2 θ)2 cos2 θ – 3 sin2 θ (cos2 θ)2

= 1 – 3 sin4 θ cos2 θ – 3 sin2 θ cos4 θ [ sin2 θ + cos2 θ = 1]

= 1 – 3 sin2 θ cos2 θ (sin2 θ + cos2 θ)

= 1 – 3 sin2 θ cos2 θ [ sin2 θ + cos2 θ = 1] ………(i)

Next, we consider

sin4 θ + cos4 θ = (sin2 θ)2 + (cos2 θ)2

Now, as (a + b)2 = a2 + b2 + 2ab

a2 + b2 = (a + b)2 – 2ab

sin4 θ + cos4 θ

= (sin2 θ)2 + (cos2 θ)2

= (sin2 θ + cos2 θ)2 – 2 sin2 θ cos2 θ

= 1 – 2 sin2 θ cos2 θ [ sin2 θ + cos2 θ = 1] ………(ii)

Now, using (i) and (ii), we have

2(sin6 θ + cos6 θ) – 3(sin4 θ + cos4 θ)

= 2(1 – 3 sin2 θ cos2 θ) – 3(1 – 2 sin2 θ cos2 θ)

= 2 – 6 sin2 θ cos2 θ – 3 + 6 sin2 θ cos2 θ

= 2 – 3 = – 1

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