Question

If A_n be the area bounded by the curve y = (tan x)ⁿ and the lines x = 0, y = 0 and x = π/4, then for x > 2
(a) A_n + A_{n −2} = $\frac{1}{n-1}$

(b) A_n + A_{n − 2} < $\frac{1}{n-1}$

(c) A_n − A_{n − 2} = $\frac{1}{n-1}$

(d) none of these

Answer 1

(a) A_n + A_{n −2} = $\frac{1}{n-1}$
$A_n=$Area bounded by the curve $y={\left\{\tan \left(x\right)\right\}}^n={\tan }^n\left(x\right)$ and the lines $x=0,y=0,$and $x=\frac{\mathrm{\pi }}{4}$.

Therefore,

$$\begin{gathered} A_n={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\tan }^n\left(x\right)\mathit{d}x \\ \Rightarrow A_{n-2}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\tan }^{n-2}\left(x\right)\mathit{d}x \end{gathered}$$
Consider, $A_n={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\tan }^n\left(x\right)\mathit{d}x$
$$\begin{gathered} \Rightarrow A_n={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\left\{{\tan }^{n-2}\left(x\right)\right\}\left\{{\tan }^2\left(x\right)\right\}\mathit{d}x \\ \Rightarrow A_n={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\left\{{\tan }^{n-2}\left(x\right)\right\}\left\{{\sec }^2\left(x\right)-1\right\}\mathit{d}x \\ \Rightarrow A_n={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\left\{{\tan }^{n-2}\left(x\right){\sec }^2\left(x\right)-{\tan }^{n-2}\left(x\right)\right\}\mathit{d}x \\ \Rightarrow A_n={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\left\{{\tan }^{n-2}\left(x\right){\sec }^2\left(x\right)\right\}\mathit{d}x-{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\tan }^{n-2}\left(x\right)\mathit{d}x \end{gathered}$$
$\Rightarrow A_n+A_{n-2}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\tan }^{n-2}\left(x\right){\sec }^2\left(x\right)\mathit{d}x$

Now, $A_n+A_{n-2}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\tan }^{n-2}\left(x\right){\sec }^2\left(x\right)\mathit{d}x$

$$\begin{gathered} \mathrm{Let}u=\tan \left(x\right) \\ \Rightarrow \mathit{d}u={\sec }^2\left(x\right)dx \end{gathered}$$
Also, when $x=0,u=0$ and when $x=\frac{\mathrm{\pi }}{4},u=1$

Therefore,

$A_n+A_{n-2}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\tan }^{n-2}\left(x\right){\sec }^2\left(x\right)\mathit{d}x$
$$\begin{gathered} ={\int }_0^1\left(u^{n-2}\right)du \\ ={\left[\frac{u^{n-1}}{n-1}\right]}_0^1 \\ =\left[\frac{1}{n-1}-0\right]=\frac{1}{n-1} \end{gathered}$$