Inverse Circular Functions Ques 28
- If $x, y$ and $z$ are in AP and $\tan ^{-1} x, \tan ^{-1} y$ and $\tan ^{-1} z$ are also in AP, then
(2013 Main)
(a) $x=y=z$
(b) $2 x=3 y=6 z$
(c) $6 x=3 y=2 z$
(d) $6 x=4 y=3 z$
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Answer:
Correct Answer: 28.(a)
Solution:
Formula:
Identities of Addition and Substraction:
- Since, $x, y$ and $z$ are in an AP.
$\therefore \quad 2 y=x+z$
Also, $\tan ^{-1} x, \tan ^{-1} y$ and $\tan ^{-1} z$ are in an AP.
$ \begin{array}{ll} \therefore & 2 \tan ^{-1} y=\tan ^{-1} x+\tan ^{-1}(z) \\ \Rightarrow & \tan ^{-1} (\frac{2 y}{1-y^{2}})=\tan ^{-1} (\frac{x+z}{1-x z}) \\ \Rightarrow & \frac{x+z}{1-y^{2}}=\frac{x+z}{1-x z} \Rightarrow y^{2}=x z \end{array} $
Since $x, y$ and $z$ are in an AP as well as in a GP.
$ \therefore \quad x=y=z $
BETA
