If -x-cos-y-cos-2 -y-z-cos-2 - then x-y-z-

The correct option is C 0We have x/cos θ=y/cos(θ 2π/3)=z/cos(θ+2π/3)=k ⇒ x=kcos θ, y=kcos(θ 2π/3), z=kcos(θ+2π/3) ⇒ x+y+z=k [cos θ+cos(θ 2π/3)+ cos(θ+2π/3)] =k ...

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Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

If .x/cosθ=y/...

Question

If $\frac{x}{c o s θ} = \frac{y}{c o s (θ - \frac{2 π}{3})} = \frac{z}{c o s (θ + \frac{2 π}{3})},$ then x + y + z=

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Solution

We have $\frac{x}{c o s θ} = \frac{y}{c o s (θ - \frac{2 π}{3})} = \frac{z}{c o s (θ + \frac{2 π}{3})} = k$
$\Rightarrow x = k c o s θ, y = k c o s (θ - \frac{2 π}{3}), z = k c o s (θ + \frac{2 π}{3})$
$\Rightarrow x + y + z = k [c o s θ + c o s (θ - \frac{2 π}{3}) + c o s (θ + \frac{2 π}{3})]$
$= k [(0) = 0$
$\Rightarrow x + y + z = 0.$

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If xcos θ=ycos(θ−2π3)=zcos(θ+2π3), then x + y + z=

We have xcos θ=ycos(θ−2π3)=zcos(θ+2π3)=k ⇒x=kcos θ,y=kcos(θ−2π3),z=kcos(θ+2π3) ⇒x+y+z=k[cos θ+cos(θ−2π3)+cos(θ+2π3)] =k[(0)=0 ⇒x+y+z=0.

If $\frac{x}{c o s θ} = \frac{y}{c o s (θ - \frac{2 π}{3})} = \frac{z}{c o s (θ + \frac{2 π}{3})},$ then x + y + z=