Let An be the area bounded by the curve y=(tanx)n and the lines x=0,y=0, and x=4π˙ Prove that for n>2,An+An−2=n−11 and deduce 2n+21<an<1(2n−2)

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Let {A}_{{n}} be the area bounded by the curve {y}={({{{x}}})}

Question

Let $A_{n}$ be the area bounded by the curve $y = (tan x)^{n}$ and the lines $x = 0, y = 0,$ and $x = \frac{π}{4} ˙$ Prove that for $n > 2, A_{n} + A_{n - 2} = \frac{1}{n - 1}$ and deduce $\frac{1}{2 n + 2} < a_{n} < 1 (2 n - 2)$

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Let

A_{n}

be the area bounded by the curve

y = (tan x)^{n}

and the lines

x = 0, y = 0,

and

x = \frac{π}{4} ˙

Prove that for

n > 2, A_{n} + A_{n - 2} = \frac{1}{n - 1}

and deduce

\frac{1}{2 n + 2} < a_{n} < 1 (2 n - 2)

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Question Text

Let

A_{n}

be the area bounded by the curve

y = (tan x)^{n}

and the lines

x = 0, y = 0,

and

x = \frac{π}{4} ˙

Prove that for

n > 2, A_{n} + A_{n - 2} = \frac{1}{n - 1}

and deduce

\frac{1}{2 n + 2} < a_{n} < 1 (2 n - 2)