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प्रश्न
If x sin3θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ , then show that x2 + y2 = 1.
उत्तर
Given: x sin3 θ + y cos3 θ = sin θ. cos θ
⇒ (x sin θ) sin2θ + (y cos θ) cos2θ = sin θ. cos θ
⇒ (x sin θ) sin2θ + (x sin θ) cos2θ = sin θ. cos θ .....(∵ y cos θ = x sin θ)
⇒ x sin θ ( sin2θ + cos2θ ) = sin θ. cos θ
⇒ x sin θ = sin θ. cos θ
⇒ x = cos θ ....(1)
Again x sin θ = y cos θ
⇒ cos θ sin θ = y cos θ
⇒ y = sin θ .....(2)
Squaring and adding (1) and (2), we get the required result.
Hence proved.
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संबंधित प्रश्न
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( 1 + cotθ - cosecθ) ( 1 + tanθ + secθ)
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Without using trigonometric identity , show that :
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`sec70^circ sin20^circ - cos20^circ cosec70^circ = 0`
If tan θ = `13/12`, then cot θ = ?
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
Find the value of sin2θ + cos2θ
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`