Question

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

• A. $\frac{(n+1{)}^2}{2n}$
• B. $\frac{(n+1{)}^2}{n}$
• C. $\frac{(n+1{)}^2}{2}$
• D. $\frac{(n-1{)}^2}{4n}$

Answer 1

Answer: (D)

$\frac{(n-1{)}^2}{4n}$

$\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}$

we know, $\tan x=n\tan y$

$\tan (x-y)=\frac{\frac{\tan x}{\tan y}-1}{\frac{1}{\tan y}+\tan x}$

$\tan (x-y)=\frac{n-1}{\frac{1}{\tan y}+n\tan y}$

for $\tan (x-y)$ to be max

$\frac{1}{\tan y}+n\tan y$ should be min

$A.M\ge G.M$

$\frac{\frac{1}{\tan y}+n\tan y}{2}\ge \sqrt{\frac{1}{\tan y}\times n\tan y}$

$\frac{1}{\tan y}+n\tan y\ge 2\sqrt{n}$

$\tan (x-y)=\frac{n-1}{2\sqrt{n}}$

squaring both sides

${\tan }^2(x-y)=\frac{n^2+1-2n}{4n}$

$=\frac{{(n-1)}^2}{4n}$