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प्रश्न
2 (sin6 θ + cos6 θ) − 3 (sin4 θ + cos4 θ) is equal to
पर्याय
0
1
−1
None of these
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उत्तर
The given expression is `2(sin^6θ+cos^6θ)-3(sin^4θ+cos^4θ)`
Simplifying the given expression, we have
`2(sinθ+cos^6θ)-3(sin^4θ+cos^4θ)`
= `2sin^6θ+2cos^6θ-3sin^4θ-3cos^4θ`
=`(2 sin^6 θ-3sin^4θ)+(2 cos^6-3 cos^4θ)`
=`sin^4θ(2sin^2θ-3)+cos^4θ(2 cos^2θ-3)`
`=sin^4θ{2(1-cos^2)-3}+cos^4θ{2(1-sin^2 θ)-3)`
`= sin^4θ(2-2cos^2θ-3)+cos^4θ(2-2sin^2 θ-3) `
`=sin^4θ(-1-2cos^θ)+cos^4θ(1-2sin^2θ)`
`= -sin^4θ-2 sin^4θ cos^2θ-cos^4θ-2cos^4 θ sin^2θ`
`=sin^4θ-cos^4θ-2 cos^4 θ sin^2θ-2 sin^4 θcos^2θ`
`=-sin^4θ-cos^4θ-2cos^2θ sin^2(cos^2+sin^2θ)`
`=-sin^4θ-cos^4θ-2cos^2θsin^2θ(1)`
`=-sin^4θ-cos^4θ-2cos^2sin^2θ`
`=(sin^4θ+cos^4 θ+2 cos^2 θ sin^2 θ)`
`=-{(sin^2θ)^2+(cos^2θ)^2+2 sin^2 θ cos^2θ}`
` =-(sin^2θ+cos^2θ)^2`
`=-(1)^2`
`=-1`
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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`
