Given, tan−1x,tan−1y and tan−1z are in A.P., so
2tan−1y=tan−1x+tan−1z
⇒tan−1y−tan−1x=tan−1z−tan−1y
1+xyy−x=1+yzz−y...(i)
Given x,y,z are in A.P.
i.e., 2y=x+z⇒y−x=z−y
Hence, from (i)
1+xy=1+yz
xy=yz either y=0 (rejected) or x=z
Hence, x=y=z.
If x,y,z are positive numbers in A.P. and tan−1x, tan−1y and tan−1z are also in A.P., then which of the following is correct.
Held on 7 Apr 2013 · Verified 6 Jul 2026.
6x=3y=2z
6x=4y=3z
x=y=z
2x=3y=6z
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