Question

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

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

Answer 1

The correct option is B $\frac{(n+1{)}^2}{4n}$

Given:
$\tan x=n\tan y$
We know that,
$\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}\Rightarrow \tan (x-y)=\frac{n\tan y-\tan y}{1+n{\tan }^{2}y}\Rightarrow \tan (x-y)=\frac{(n-1)\tan y}{1+n{\tan }^{2}y}\Rightarrow {\sec }^2(x-y)-1=\frac{(n-1{)}^2{\tan }^{2}y}{(1+n{\tan }^{2}y{)}^2}\cdots \left(1\right)$

Now, using $A.M.\ge G.M.$ for $1,n{\tan }^{2}y$, we get
$\frac{1+n{\tan }^{2}y}{2}\ge \sqrt{n}\tan y$
Squaring both sides,
$\Rightarrow \frac{(1+n{\tan }^{2}y{)}^2}{4}\ge n{\tan }^{2}y\Rightarrow (1+n{\tan }^{2}y{)}^2\ge 4n{\tan }^{2}y\Rightarrow \frac{1}{4n}\ge \frac{{\tan }^{2}y}{(1+n{\tan }^{2}y{)}^2}\Rightarrow \frac{(n-1{)}^2}{4n}\ge \frac{(n-1{)}^2{\tan }^{2}y}{(1+n{\tan }^{2}y{)}^2}$
Using equation $\left(1\right)$, we get
$\Rightarrow \frac{(n-1{)}^2}{4n}\ge {\sec }^2(x-y)-1\Rightarrow 1+\frac{(n-1{)}^2}{4n}\ge {\sec }^2(x-y)\therefore \frac{(n+1{)}^2}{4n}\ge {\sec }^2(x-y)$

Hence, the minimum value of ${\sec }^2(x-y)$ is $\frac{(n+1{)}^2}{4n}$