Inverse Trigonometric FunctionsmediumFree

Inverse Trigonometric Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

\cot^{-1}\left(4+\dfrac{2}{4}\right)+\cot^{-1}\left(4+\dfrac{6}{4}\right)+\cot^{-1}\left(4+\dfrac{12}{4}\right)+\ldots\infty=\tan^{-1}\left(\dfrac{a}{b}\right)

where

a

and

b

are coprime, then find the value of

(a^{3}+b^{3})

Solution

Answer: 65

Step 1: Observe the pattern:

2,6,12,\ldots

has general term

r(r+1)

. So the

r

-th term is:

T_{r}=\cot^{-1}\left(4+\dfrac{r(r+1)}{4}\right)=\cot^{-1}\left(\dfrac{16+r(r+1)}{4}\right)=\tan^{-1}\dfrac{4}{16+r(r+1)}

Step 2: Manipulate the argument:

\dfrac{4}{16+r(r+1)}=\dfrac{\frac{r+1}{4}-\frac{r}{4}}{1+\frac{r}{4}\cdot\frac{r+1}{4}}

Step 3: So:

T_{r}=\tan^{-1}\dfrac{r+1}{4}-\tan^{-1}\dfrac{r}{4}

Step 4: Telescoping sum:

S_{n}=\sum_{r=1}^{n}T_{r}=\tan^{-1}\dfrac{n+1}{4}-\tan^{-1}\dfrac{1}{4}

Step 5: Taking

n\to\infty

S_{\infty}=\dfrac{\pi}{2}-\tan^{-1}\dfrac{1}{4}=\cot^{-1}\dfrac{1}{4}=\tan^{-1}4=\tan^{-1}\dfrac{a}{b}

Step 6: So

a=4

and

b=1

. Hence:

a^{3}+b^{3}=64+1=65

Answer:

65

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