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