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Question
`sin^6 theta + cos^6 theta =1 -3 sin^2 theta cos^2 theta`
Solution
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
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
