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प्रश्न
Prove that sin6θ + cos6θ = 1 – 3 sin2θ. cos2θ.
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उत्तर
LHS = sin6θ + cos6θ
= (sin2θ)3 + (cos2θ)3
= (sin2θ + cos2θ) (sin4θ + cos4θ - sin2θ⋅cos2θ)
= (1)[(sin2θ + cos2θ)2 - 2sin2θ⋅cos2θ - sin2θ⋅cos2θ]
= (1)[(1)2 - 3sin2θ⋅cos2θ]
= 1 - 3sin2θ ⋅ cos2θ
= RHS
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Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
