If x cos -y cos -2 -3 -z cos -4 -3 - then the value of 1-x-1-y-1-z is equal to

We have x cos θ=y cos ( θ+2 π/3 )=z cos ( θ+4 π/3 )=k ⇒ cos θ=k/x, cos ( θ+2 π/3 )=k/y and cos ( θ+4 π/3 )=k/z Hence k/x+k/y+k/z=cos θ+cos ( θ+2 π/3 )+cos ( ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

If x cos θ=y ...

Question

$I f x c o s θ = y c o s (θ + \frac{2 π}{3}) = z c o s (θ + \frac{4 π}{3}), t h e n t h e v a l u e o f \frac{1}{x} + \frac{1}{y} + \frac{1}{z} i s e q u a l t o$

$1$

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$2$

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$0$

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$3 c o s θ$

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Solution

The correct option is C
$0$

We have

$x c o s θ = y c o s (θ + \frac{2 π}{3}) = z c o s (θ + \frac{4 π}{3}) = k \Rightarrow c o s θ = \frac{k}{x}, c o s (θ + \frac{2 π}{3}) = \frac{k}{y} a n d c o s (θ + \frac{4 π}{3}) = \frac{k}{z} H e n c e \frac{k}{x} + \frac{k}{y} + \frac{k}{z} = c o s θ + c o s (θ + \frac{2 π}{3}) + c o s (θ + \frac{4 π}{3})$
$= c o s θ + c o s (\frac{π}{3} - θ) - c o s (\frac{π}{3} + θ) = c o s θ - 2 c o s \frac{π}{3} c o s θ = 0$

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Similar questions

Q. 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=

Q. If

{cos}^{- 1} x + {cos}^{- 1} y + {cos}^{- 1} z = 3 π,

then the value of

x^{3} + y^{3} + z^{3} - 3 x y z

$I f x c o s θ = y c o s (θ + \frac{2 π}{3}) = z c o s (θ + \frac{4 π}{3}), t h e n t h e v a l u e o f \frac{1}{x} + \frac{1}{y} + \frac{1}{z} i s e q u a l t o$

Q. If

a = sin θ

b = sin (θ + 2 π / 3)

c = sin (θ + 4 π / 3)

x = cos θ

y = cos (θ + 2 π / 3)

z = cos (θ + 4 π / 3)

, then value of

Δ = ∣ ∣ ∣ ∣ \begin{matrix} a & b & c x & y & z b c & c a & a b \end{matrix} ∣ ∣ ∣ ∣

If {cos}^{- 1} x + {cos}^{- 1} y + {cos}^{- 1} z = 3 π,

then the value of

x y + y z + z x

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If x cos θ=y cos (θ+2π3)=z cos(θ+4π3), then the value of 1x+1y+1z is equal to

The correct option is C 0 We have x cos θ=y cos (θ+2π3)=z cos(θ+4π3)=k⇒cos θ=kx, cos(θ+2π3)=kyand cos(θ+4π3)=kzHence kx+ky+kz=cos θ+cos(θ+2π3)+cos(θ+4π3) =cos θ+cos(π3−θ)−cos(π3+θ)=cos θ−2 cosπ3cos θ=0

$I f x c o s θ = y c o s (θ + \frac{2 π}{3}) = z c o s (θ + \frac{4 π}{3}), t h e n t h e v a l u e o f \frac{1}{x} + \frac{1}{y} + \frac{1}{z} i s e q u a l t o$