Question

If x y z are in gp and tan^-1(x) tan^-1(y) tan^-1(z) are in ap then?

Answer 1

Given:

x y z are in GP and

tan^-1(x) tan^-1(y) tan^-1(z) are in AP.

To Find:

The expression which satisfies the above given conditions.

Solution:

• x, y, z are in GP = > y² = xz - (a )

• tan^-1(x) tan^-1(y) tan^-1(z) are in AP =  >
• $tan^{-1}$x + $tan^{-1}$ z = 2$tan^{-1}$y
• $tan^{-1} ( \frac{x+ z}{1 - xz} )$ = 2$tan^{-1}$y
• $\frac{x+z}{1-xz}$ = tan ( 2$tan^{-1}$y )

We know ,

• tan 2x = $\frac{2tanx}{1 - tan^{2}x }$
• Therefore, tan(2$tan^{-1}$y) =  $\frac{2tan(tan^{-1}y) }{1 - (tan (tan^{-1} y))^{2} }$  =  2y/1-y²
• $\frac{x + z}{1-xz}$ = 2y/1-y²

We already so that xz = y² .

Therefore,

• 1 -xz = 1 -y²

Applying this , we get.

• x + z = 2y

If x y z are in gp and tan^-1(x) tan^-1(y) tan^-1(z) are in ap then x + z =2y