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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.
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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.
संबंधित प्रश्न
`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))=0`
Prove that cosec2 (90° - θ) + cot2 (90° - θ) = 1 + 2 tan2 θ.
Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.
Prove the following identities:
`(1 - tan^2 θ)/(cot^2 θ - 1) = tan^2 θ`.
Prove the following identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
sin2θ + sin2(90 – θ) = ?
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
If tan θ – sin2θ = cos2θ, then show that `sin^2θ = 1/2`.
If sin A = `1/2`, then the value of sec A is ______.
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
