Question

If tan x =n tany, $n\in R^{+}$, then maximum value of $se{c}^2(x-y)=$___

• 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

The correct option is D

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

tan x=n tan y
cos (x-y)=cosx cosy +sin x sin y
=cos x cos y (1+ tan x tan y)
$=cosxcosy(1+nta{n}^{2}y)se{c}^2(x-y)=\frac{se{c}^{2}xse{c}^{2}y}{(1+nta{n}^{2}y{)}^2}=\frac{(1+ta{n}^{2}x)(1+ta{n}^{2}y)}{(1+nta{n}^{2}y{)}^2}=\frac{(1+n^{2}ta{n}^{2}y)(1+ta{n}^{2}y)}{(1+nta{n}^{2}y{)}^2}=1+\frac{(n-1{)}^{2}ta{n}^{2}y}{(1+nta{n}^{2}y{)}^2}$
Now, ${\left(\frac{1+nta{n}^{2}y}{2}\right)}^2\ge nta{n}^{2}y$
$\Rightarrow \frac{ta{n}^{2}y}{(1+ta{n}^{2}y{)}^2}\le \frac{1}{4n}\Rightarrow se{c}^2(x-y)\le 1+\frac{(n-1{)}^2}{4n}=\frac{(n+1{)}^2}{4n}.$