If x sin-y sin-2 -3-z sin-4 -3- then the value of x y-y z-z x is

Let xsinθ=ysin(θ+2π3)=zsin(θ+4π3)=k ⇒ xsinθ=ysin(θ+120^∘)=zsin(θ+240^∘)=k x =ksinθ y = ksin(θ+120^∘) = ksin(180^∘ (θ+120^∘)) ⇒ y = nbsp;ksin(60^∘ θ) z =k ...

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Byju's Answer

Standard XII

Mathematics

Tangent, Cotangent, Secant, Cosecant in Terms of Sine and Cosine

If x sinθ=y s...

Question

If $x sin θ = y sin (θ + \frac{2 π}{3}) = z sin (θ + \frac{4 π}{3})$ , then the value of $x y + y z + z x$ is

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Solution

Let $x sin θ = y sin (θ + \frac{2 π}{3}) = z sin (θ + \frac{4 π}{3}) = k \Rightarrow x sin θ = y sin (θ + 120^{\circ}) = z sin (θ + 240^{\circ}) = k x = \frac{k}{sin θ} y = \frac{k}{sin (θ + 120^{\circ})} = \frac{k}{sin (180^{\circ} - (θ + 120^{\circ}))} \Rightarrow y = \frac{k}{sin (60^{\circ} - θ)} z = \frac{k}{sin (θ + 240^{\circ})} = \frac{k}{sin (180^{\circ} + (θ + 60^{\circ}))} \Rightarrow z = - \frac{k}{sin (60^{\circ} + θ)}$

So,
$x y + y z + z x = k^{2} [\frac{1}{sin θ sin (60^{\circ} - θ)} - \frac{1}{sin (60^{\circ} + θ) sin (60^{\circ} - θ)} - \frac{1}{sin θ sin (60^{\circ} + θ)}] = k^{2} [\frac{sin (60^{\circ} + θ) - sin θ - sin (60^{\circ} - θ)}{sin θ sin (60^{\circ} - θ) sin (60^{\circ} + θ)}] = 4 k^{2} [\frac{2 cos 60^{\circ} sin θ - sin θ}{sin 3 θ}] = 4 k^{2} [\frac{sin θ - sin θ}{sin 3 θ}] = 0$

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Similar questions

Q. If

x sin θ = y sin (θ + \frac{2 π}{3}) = z sin (θ + \frac{4 π}{3})

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Q. If

x cos θ = y cos (θ + \frac{2 π}{3}) = z cos (θ + \frac{4 π}{3})

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x y + y z + z x =

If $x c o s θ = y c o s (θ + \frac{2 π}{3}) = z c o s (θ + \frac{4 π}{3})$ , prove that $x y + y z + z x = 0$

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x cos θ = y cos (θ + \frac{2 π}{3}) = z cos (θ + \frac{4 π}{3})

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x y + y z + z x = 0

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If xsinθ=ysin(θ+2π3)=zsin(θ+4π3), then the value of xy+yz+zx is

If $x sin θ = y sin (θ + \frac{2 π}{3}) = z sin (θ + \frac{4 π}{3})$ , then the value of $x y + y z + z x$ is