Question

# If $$\tan x =n\tan y, n\in R^+$$, then the maximum value of $$\tan^2(x-y)$$ is equal to

A
(n+1)22n
B
(n+1)2n
C
(n+1)22
D
(n1)24n

Solution

## The correct option is C $$\dfrac{(n-1)^2}{4n}$$$$\tan(x-y)=\cfrac { \tan x-\tan y }{ 1+\tan x\tan y }$$we know, $$\tan x=n\tan y$$$$\tan(x-y)=\cfrac { \cfrac { \tan x }{ \tan y } -1 }{ \cfrac { 1 }{ \tan y } +\tan x }$$$$\tan(x-y)=\cfrac { n-1 }{ \cfrac { 1 }{ \tan y } +n\tan y }$$for $$\tan(x-y)$$ to be max$$\cfrac { 1 }{ \tan y } +n\tan y$$ should be min$$A.M\ge G.M$$$$\cfrac { \cfrac { 1 }{ \tan y } +n\tan y }{ 2 } \ge \sqrt { \cfrac { 1 }{ \tan y } \times n\tan y }$$$$\cfrac { 1 }{ \tan y } +n\tan y \ge 2\sqrt{n}$$$$\tan(x-y)=\cfrac { n-1 }{ 2\sqrt { n } }$$squaring both sides$${ \tan }^{ 2 }(x-y)=\cfrac { { n }^{ 2 }+1-2n }{ 4n }$$$$=\cfrac { { (n-1) }^{ 2 } }{ 4n }$$Maths

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