If tanx =ntany, n e R⁺, then the maximum value of sec²(x-y) is equal to-

Solution: (d); Say, $z=\sec ^{2}(x-y)=1+\tan ^{2}(x-y)=1+\left(\frac{\tan x-\tan y}{1+\tan x \cdot \tan y}\right)^{2}$

$\begin{array}{l} =1+\left(\frac{n \tan y-\tan y}{1+n \cdot \tan y \cdot \tan y}\right)^{2}[\because \tan x=n \cdot \tan y] \\ \Rightarrow z=1+\left(\frac{(n-1) \tan y}{1+n \cdot \tan ^{2} y}\right)^{2}=1+(n-1)^{2}\left(\frac{\tan y}{1+n \cdot \tan ^{2} y}\right)^{2} \Rightarrow z=1+(n-1)^{2}\left(\frac{t}{1+n t^{2}}\right)^{2}[\text { Putting, tan } y=z] \end{array}$

Now, for maximum and minimum, $\frac{\mathrm{dz}}{\mathrm{dt}}=0 \Rightarrow 0+2(\mathrm{n}-1)^{2}\left(\frac{\mathrm{t}}{1+\mathrm{nt}^{2}}\right) \cdot \frac{\left(1+\mathrm{nt}^{2}\right) \cdot 1-\mathrm{t} \cdot(0+2 \mathrm{nt})}{\left(1+\mathrm{nt}^{2}\right)^{2}}=0$

$\begin{array}{l} \Rightarrow \mathrm{t}\left(1+\mathrm{nt}^{2}-2 \mathrm{nt}^{2}\right)=0 \Rightarrow \mathrm{t}\left(1-\mathrm{nt}^{2}\right)=0[\because \mathrm{t} \neq 0] \Rightarrow 1-\mathrm{nt}^{2}=0 \therefore \mathrm{t}^{2}=\frac{1}{\mathrm{n}} \\ \therefore \mathrm{z}_{\max }=1+(\mathrm{n}-1)^{2} \frac{\mathrm{t}^{2}}{\left(1+n \mathrm{t}^{2}\right)^{2}}=1+(\mathrm{n}-1)^{2} \times \frac{\frac{1}{n}}{\left(1+\mathrm{n} \cdot \frac{1}{\mathrm{n}}\right)^{2}} \\ =1+\frac{(\mathrm{n}-1)^{2}}{4 \mathrm{n}}=\frac{4 \mathrm{n}+(\mathrm{n}-1)^{2}}{4 \mathrm{n}}=\frac{(\mathrm{n}-1)^{2}+4 \cdot \mathrm{n} \cdot 1}{4 \mathrm{n}}=\frac{(\mathrm{n}+1)^{2}}{4 \mathrm{n}} \text { (Ans.) } \end{array}$