Question

Let $\alpha ,\beta ,\gamma$ be in AP and $x,y,z$ be in GP. If $\tan \alpha =x,\tan \beta =y$ and $\tan \gamma =z,$ then

• A. $x=y=z$
• B. $xz=1$
• C. $x=y=z\ne 1$
• D. $x,y,z$ are in AP

Answer 1

Answer: (C), (D)

The correct options are
C $x=y=z\ne 1$
D $x,y,z$ are in AP

$x,y,z$ are in GP.
$\Rightarrow y^2=xz$

$\alpha ,\beta ,\gamma$ are in AP.
$\Rightarrow 2\beta =\alpha +\gamma$
$\Rightarrow \tan \left(2\beta \right)=\tan (\alpha +\gamma )$
$\Rightarrow \frac{2\tan \beta }{1-{\tan }^2\beta }=\frac{\tan \alpha +\tan \gamma }{1-\tan \alpha \tan \gamma }$
$\Rightarrow \frac{2y}{1-y^2}=\frac{x+z}{1-xz}$
$\Rightarrow 2y=x+z$ and $y\ne 1$
$\Rightarrow x,y,z$ are in AP.
Since, $x,y,z$ are in AP and GP,
$x,y,z$ is a constant sequence.