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

विकल्प

• A_n + A_n −2 = \[\frac{1}{n - 1}\]

• A_n + A_{n − 2} < \[\frac{1}{n - 1}\]

• A_n − A_{n − 2} = \[\frac{1}{n - 1}\]

• none of these

Answer 1

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 \[x = \frac{\pi}{4}\]
Therefore,

\[A_n = \int_0^\frac{\pi}{4} \tan^n \left( x \right) d x\]
\[ \Rightarrow A_{n - 2} = \int_0^\frac{\pi}{4} \tan^{n - 2} \left( x \right) d x\]
Consider, "
\[A_n = \int_0^\frac{\pi}{4} \tan^n \left( x \right) d x\]
\[\Rightarrow A_n = \int_0^\frac{\pi}{4} \left\{ \tan^{n - 2} \left( x \right) \right\}\left\{ \tan^2 \left( x \right) \right\} d x\]
\[ \Rightarrow A_n = \int_0^\frac{\pi}{4} \left\{ \tan^{n - 2} \left( x \right) \right\}\left\{ \sec^2 \left( x \right) - 1 \right\} d x\]
\[ \Rightarrow A_n = \int_0^\frac{\pi}{4} \left\{ \tan^{n - 2} \left( x \right) \sec^2 \left( x \right) - \tan^{n - 2} \left( x \right) \right\} d x\]
\[ \Rightarrow A_n = \int_0^\frac{\pi}{4} \left\{ \tan^{n - 2} \left( x \right) \sec^2 \left( x \right) \right\} d x - \int_0^\frac{\pi}{4} \tan^{n - 2} \left( x \right) d x\]

\[\Rightarrow A_n + A_{n - 2} = \int_0^\frac{\pi}{4} \tan^{n - 2} \left( x \right) \sec^2 \left( x \right) d x\]
Now, 

\[A_n + A_{n - 2} = \int_0^\frac{\pi}{4} \tan^{n - 2} \left( x \right) \sec^2 \left( x \right) d x\]
\[\text{Let }u = \tan\left( x \right)\]
\[ \Rightarrow d u = \sec^2 \left( x \right)dx\]
Also, when \[x = 0, u = 0\] and when \[x = \frac{\pi}{4}, u = 1\]
Therefore,

\[A_n + A_{n - 2} = \int_0^\frac{\pi}{4} \tan^{n - 2} \left( x \right) \sec^2 \left( x \right) d x\]
\[= \int_0^1 \left( u^{n - 2} \right) d u\]
\[ = \left[ \frac{u^{n - 1}}{n - 1} \right]_0^1 \]
\[ = \left[ \frac{1}{n - 1} - 0 \right] = \frac{1}{n - 1}\]