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Question Number 8224 by trapti rathaur@ gmail.com last updated on 03/Oct/16

i f \emptyset l i e s b e t w e e n - \frac{π}{4} a n d \frac{π}{4} t h e n p r o v e t h a t

\emptyset^{2} = \tan^{2} \emptyset - (1 + \frac{1}{3}) \frac{\tan^{4} \emptyset}{2} + (1 + \frac{1}{3} + \frac{1}{5}) \frac{\tan^{6} \emptyset}{3} + - - - - -

- - - t o \infty t e r m s

Commented by Yozzias last updated on 03/Oct/16

RTP : If b \in (\frac{- π}{4}, \frac{π}{4}) \Rightarrow b^{2} = \underset{r = 1}{\sum^{\infty}} {(- 1)}^{r - 1} \frac{1}{r} {\underset{k = 1}{\sum^{r}} \frac{1}{2 k - 1}} \tan^{2 r} b .

or b^{2} = \lim_{n \to \infty} [\underset{r = 1}{\sum^{n}} {(- 1)}^{r - 1} \frac{1}{r} {\underset{k = 1}{\sum^{r}} \frac{1}{2 k - 1}} \tan^{2 r} b]

Commented by trapti rathaur@ gmail.com last updated on 03/Oct/16

H i n t - \tan^{- 1} x + \tan^{- 1} y + \tan^{- 1} z = \tan^{- 1} \frac{x + y + z - x y z}{1 - x y - y z - z x}

Commented by trapti rathaur@ gmail.com last updated on 03/Oct/16

i n t h e q u e s t i o n g i v e n a b o v e h i n t

s o h o w t o s o l v e t h i s q u e s t i o n u s i n g h i n t .

Answered by prakash jain last updated on 03/Oct/16

Taylor series expansion for \tan^{- 1} x

\tan^{- 1} x = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} + - \dots (for - 1 < x < 1)

{(\tan^{- 1} x)}^{2} = (x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - + \dots) (x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - + \dots)

w h e n n i s e v e n c o e f f i c i e n t o f x^{n} i s o b t a i n e d

b y m u l t i p l y i n g c o e e f i c i e n t s

f o r e x a m p l e c o e f f i c i e n t o f x^{6} = a_{1} a_{5} + a_{3} a_{3} + a_{5} a_{1}

w h e r e a_{n} i s c o e f f i i e n t o f x^{n} i n \tan^{- 1} x

c_{n} = coefficient of x^{n} = {\begin{cases} \underset{k = 1}{\sum^{n / 2}} \frac{1}{(2 k - 1) (2 (n - k) + 1)} & n e v e n \\ 0 & n o d d \end{cases}

\underset{k = 1}{\sum^{n / 2}} \frac{1}{(2 k - 1) (2 (n - k) + 1)}

= \underset{k = 1}{\sum^{n / 2}} \frac{1}{n} (\frac{1}{2 k - 1} + \frac{1}{2 (\frac{n}{2} - k) + 1})

= \frac{1}{n} (\underset{k = 1}{\sum^{n / 2}} \frac{1}{2 k - 1} + \underset{k = 1}{\sum^{n / 2}} \frac{1}{2 (\frac{n}{2} - k) + 1})

= \frac{1}{n} (\underset{k = 1}{\sum^{n / 2}} \frac{1}{2 k - 1} + \underset{k = 1}{\sum^{n / 2}} \frac{1}{2 k - 1})

c_{2} = \frac{2}{n} \underset{k = 1}{\sum^{n / 2}} \frac{1}{2 k - 1}

c_{n} = {\begin{cases} \frac{2}{n} \underset{k = 1}{\sum^{n / 2}} \frac{1}{2 k - 1} & n e v e n \\ 0 & n o d d \end{cases}

put

x = \tan \emptyset (- \frac{π}{4} < \emptyset < \frac{π}{4})

to get the required result