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If x,y,z are in AP and `tan^(-1)x,tan^(-1)y,tan^(-1)z` are also in AP, thenA. `2x=3y=6z`B. `6x=3y=2z`C. `6x=4y=3z`D. `x=y=z` |
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Answer» Correct Answer - D `:.` x,y,z are in AP. Let `x=y-d,z=y+d " " "…….(i)"` Also, given `tan^(-1)x,tan^(-1)y,tan^(-1)z` are in AP. `:.2tan^(-1)y=tan^(-1)x+tan^(-1)z` `implies tan^(-1)((2y)/(1-y^(2)))=tan^(-1)((x+z)/(1-xz))` `implies (2y)/(1-y^(2))=(x+z)/(1-xz) implies (2y)/(1-y^(2))=(2y)/(1-(y^(2)-d^(2)))` `implies y^(2)=y^(2)-d^(2)" " [" fromEq.(i) "]` `:. d=0` From Eq. (i), x=y and z=y `:. x=y=z` Aliter `:. x,y,z" are in AP. "" " "…….(i)"` `:. 2y=x+z" " "........(ii)"` Also, `tan^(-1)x,tan^(-1)y,tan^(-1)z` are in AP. `:.2tan^(-1)y=tan^(-1)x+tan^(-1)z` `implies tan^(-1)((2y)/(1-y^(2)))=tan^(-1)((x+z)/(1-xz))` `implies (2y)/(1-y^(2))=(x+z)/(1-xz)=(2y)/(1-xz)" "[" from Eq.(ii) "]` `implies y^(2)=zx` `:. x,y,z " are in GP. "" "".........(iii)"` From Eqs. (i) and (ii) x,y,z are in AP and also in GP, then x=y=z. |
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