Let - - - be in AP and x- y-z be in GP- If tan- -x- tan- -y and tan- -z- then

X, y,z are in GP. ⇒ y^2 =xz α, β, γ are in AP. ⇒ 2β= α + γ ⇒tan (2β) =tan (α + γ) ⇒2tanβ1 tan^2 β=tanα + tanγ1 tanαtanγ ⇒2y1 y^2=x+z1 xz ⇒ 2y=x+z and y≠ 1 ⇒ x ...

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Standard XII

Mathematics

Geometric Progression

Let α, β, γ b...

Question

Let $α, β, γ$ be in AP and $x, y, z$ be in GP. If $tan α = x, tan β = y$ and $tan γ = z,$ then

x = y = z

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x z = 1

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x = y = z \neq 1

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x, y, z

are in AP

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Solution

The correct option is D $x, y, z$ are in AP
$x, y, z$ are in GP.
$\Rightarrow y^{2} = x z$

$α, β, γ$ are in AP.
$\Rightarrow 2 β = α + γ$
$\Rightarrow tan (2 β) = tan (α + γ)$
$\Rightarrow \frac{2 tan β}{1 - {tan}^{2} β} = \frac{tan α + tan γ}{1 - tan α tan γ}$
$\Rightarrow \frac{2 y}{1 - y^{2}} = \frac{x + z}{1 - x z}$
$\Rightarrow 2 y = x + z$ and $y \neq 1$
$\Rightarrow x, y, z$ are in AP.
Since, $x, y, z$ are in AP and GP,
$x, y, z$ is a constant sequence.

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Let α,β,γ be in AP and x,y,z be in GP. If tanα=x,tanβ=y and tanγ=z, then

The correct option is D x,y,z are in APx,y,z are in GP. ⇒y2=xz α,β,γ are in AP. ⇒2β=α+γ ⇒tan(2β)=tan(α+γ) ⇒2tanβ1−tan2β=tanα+tanγ1−tanαtanγ ⇒2y1−y2=x+z1−xz ⇒2y=x+z and y≠1 ⇒x,y,z are in AP. Since, x,y,z are in AP and GP, x,y,z is a constant sequence.

Let $α, β, γ$ be in AP and $x, y, z$ be in GP. If $tan α = x, tan β = y$ and $tan γ = z,$ then