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Explanation:

Given, sinθ + cosθ = m 

Taking square boths sides we get
sin2θ + cos2θ + 2 × sinθ × cosθ = m2 
⇒ 1 + 2 × sinθ × cosθ = m2 
⇒ sinθ × cosθ = (m2 - 1)/2   ...(1)

Now we need to find the value of sin6θ + cos6θ 
= (sin2θ)3 + (cos2θ)3 
= (sin2θ + cos2θ)(sin4θ + cos4θ - sin2θ × cos2θ) 
= 1 × ((sin2θ + + cos2θ)2 - 2 × sin2θ × cos2θ - sin2θ × cos2θ) 
= ((1)2 - 3 × sin2θ × cos2θ) 
= (1 - 3 × sin2θ × cos2θ) 
= (1 - 3 × ((m2 - 1)/2)2
= (1 - 3 × (m2 - 1)2/4) 

Hence, option (d).

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