Question

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

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

Let, $z={\sec }^2(x-y)=1+{\tan }^2(x-y)$
$\Rightarrow z=1+{\left(\frac{\tan x-\tan y}{1+\tan x\tan y}\right)}^2=1+{\left(\frac{n\tan y-\tan y}{1+n{\tan }^{2}y}\right)}^2$
$\Rightarrow z=1+(n-1{)}^2{\left(\frac{\tan y}{1+n{\tan }^{2}y}\right)}^2=1+(n-1{)}^2{\left(\frac{t}{1+n{t}^2}\right)}^2$ by putting $t=\tan y$
Now for max value of $z$ we have $\frac{dz}{dt}=0$
$\Rightarrow 2(n-1{)}^2\left(\frac{t}{1+n{t}^2}\right)\left(\frac{1+n{t}^2-t\left(2nt\right)}{(1+n{t}^2{)}^2}\right)=0$
$\Rightarrow 1-n{t}^2=0\Rightarrow t^2=\frac{1}{n}$
Hence $z_{max}=1+(n-1{)}^2\left(\frac{\frac{1}{n}}{(1+n(\frac{1}{n}){)}^2}\right)=1+\frac{(n-1{)}^2}{4n}=\frac{(n+1{)}^2}{4n}$