Question

Please solve the attached question with explanation​

Answer 1

$\\ \underline{\underline{ \: \red{\Large \rm{Solution :- }}} }$

$\\ \: \: \: \: \: \: \: \: \: \: \sf \: A_n + A_n - 2 = \frac{1}{n - 1}$

In this , $\sf A_n$ is the Area bounded by the curve.

$\sf \: y = \{ \tan(x) \}^{n} = tan^{n} (x) \: and \: the \: lines \: x = 0 \: y \: = 0 \: and \: x = \dfrac{ \pi}{4}$

Therefore,

$\\ \: \: \: \sf{A_n} = \int_0^{ \frac{ \pi}{4} } \: \tan^{n} (x)dx$

$\\ \: \: \: \: \: \: \longmapsto \: \: \: \sf{A_{n - 2}} = \int_0^{ \frac{ \pi}{4} } \: \tan^{n - 2} (x)dx$

Consider,

$\\ \: \: \: \sf{A_n} = \int_0^{ \frac{ \pi}{4} } \: \tan^{n} (x)dx$

$\\ \: \: \: \: \: \: \longmapsto \: \: \: \sf{A_{n }} = \int_0^{ \frac{ \pi}{4} } \: \{ \tan^{n - 2} (x)dx \} \{ { \tan}^{2} (x) \}dx$

$\\ \: \: \: \: \: \: \longmapsto \: \: \: \sf{A_{n }} = \int_0^{ \frac{ \pi}{4} } \: \{ \tan^{n - 2} (x)dx \} \{ { \sec}^{2}(x) - 1 \}dx$

$\\ \: \: \: \: \: \: \longmapsto \: \: \: \sf{A_{n }} = \int_0^{ \frac{ \pi}{4} } \: \{ \tan^{n - 2} (x) \sec ^{2} (x) - { \tan}^{n - 2}(x) - \}dx$

$\\ \: \: \: \: \: \: \longmapsto \: \: \: \sf{A_{n }} = \int_0^{ \frac{ \pi}{4} } \: \{ \tan^{n - 2} (x) \sec ^{2} (x) \}dx \: - \int _0^{ \frac{\pi}{4} } { \tan}^{n - 2}(x) dx$

$\\ \: \: \: \: \: \: \implies \: \: \: \sf{A_{n }} + A_{n - 2} = \int_0^{ \frac{ \pi}{4} } \: \tan^{n - 2} (x) \sec ^{2} (x) dx$

Now,

$\\ \: \: \: \: \: \: \: \: \: \sf{A_{n }} + A_{n - 2} = \int_0^{ \frac{ \pi}{4} } \: \tan^{n - 2} (x) \sec ^{2} (x) dx$

Let u = tan (x)

$\\ \: \: \: \: \sf \implies \: du = \sec ^{2} (x)dx$

Also , When x = 0 , u = 0 and when $\sf x = \dfrac{\pi}{4} , u = 1$

Therefore,

$\\ \: \: \: \: \: \: \: \: \: \sf{A_{n }} + A_{n - 2} = \int_0^{ \frac{ \pi}{4} } \: \tan^{n - 2} (x) \sec ^{2} (x) dx$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \longrightarrow \: \int_0^{1} ( {u}^{n - 2} )du$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \longrightarrow \: \bigg[ \frac{ {u}^{n - 1} }{n - 1} \bigg] _0 ^{1}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \longrightarrow \: \bigg[ \frac{ 1 }{n - 1} - 0\bigg]$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \longrightarrow \: \frac{ 1 }{n - 1}$