SATHEE: Inverse Circular Functions 3 Question 8

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) $2 x = 3 y = 6 z$

(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.

$∴ 2 y = x + z$

Also, $\tan^{- 1} x, \tan^{- 1} y$ and $\tan^{- 1} z$ are in an AP.

$\begin{array}{ll} ∴ & 2 \tan^{- 1} y = \tan^{- 1} x + \tan^{- 1} (z) \\ \Rightarrow & \tan^{- 1} \frac{2 y}{1 - y^{2}} = \tan^{- 1} \frac{x + z}{1 - x z} \\ \Rightarrow & \frac{x + z}{1 - y^{2}} = \frac{x + z}{1 - x z} \Rightarrow y^{2} = x z \end{array}$

Since $x, y$ and $z$ are in an AP as well as in a GP.

$∴ x = y = z$

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

Answer:

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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

Answer:

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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