Inverse Circular Functions 3 Question 8
8. 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: 8. (a)
Solution:
- 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.
$$ \therefore \quad x=y=z $$
