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2025-07-04

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Sri Lanka A need 225 runs
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2025-07-03

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2025-07-02

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2025-07-03
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2025-07-02
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2025-07-01
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2025-07-01

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2025-07-01

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2025-06-30
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2025-06-30
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2025-06-29
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2025-06-29

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2025-06-29

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2025-06-29

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2025-06-29

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2025-06-29

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2025-06-29

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2025-06-29

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2025-06-29

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2025-06-29

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2025-06-29

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2025-06-29

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2025-06-29
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2025-06-28
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2025-06-28

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2025-06-28

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2025-06-28

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2025-06-28

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2025-06-28
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2025-06-28

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2025-06-28

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2025-06-27

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Romania won by 7 wkts
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2025-06-27

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2025-06-27
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Widget
Thursday, August 25, 2011
Arctangent series of Madhava, Gregory and Liebniz
The inverse tangent series of Madhava is given in verse 2.206 – 2.209 in Yukti-dipika commentary (Tantrasamgraha-vyakhya) by Sankara Variar . It is also given by Jyeshtadeva in Yuktibhasha and a translation of the verses is given below.
Now, by just the same argument, the determination of the arc of a desired sine can be (made). That is as follows: The first result is the product of the desired sine and the radius divided by the cosine of the arc. When one has made the square of the sine the multiplier and the square of the cosine the divisor, now a group of results is to be determined from the (previous) results beginning from the first. When these are divided in order by the odd numbers 1, 3, and so forth, and when one has subtracted the sum of the even(-numbered) results from the sum of the odd (ones), that should be the arc. Here the smaller of the sine and cosine is required to be considered as the desired (sine). Otherwise, there would be no termination of results even if repeatedly (computed).
Rendering in modern notations
Let s be the arc of the desired sine, bhujajya, y. Let r be the radius and x be the cosine (kotijya).
The first result is y.r/x
From the divisor and multiplier y^2/x^2
From the group of results y.r/x.y^2/x^2, y.r/x. y^2/x^2.y^2/x^2
Divide in order by number 1,3 etc
1 y.r/1x, 1y.r/3x y^2/x^2, 1y.r/5x.y^2/x^2.Y^2/x^2
a = (Sum of odd numbered results) 1 y.r/1x + 1y.r/5x.y^2/x^2.y^2/x^2+......
b= ( Sum of even numbered results) 1y.r/3x.y^2/x^2 + 1 y.r/7x.y^2/x^2.y^2/x^.y^2/x^2+.....
The arc is now given by
s = a - b
Transformation to current notation
If x is the angle subtended by the arc s at the Center of the Circle, then s = rx and kotijya = r cos x and bhujajya = r sin x. And sparshajya = tan x
Simplifying we get
x = tan x - tan^3x/'3 + tan^5x/5 - tan^7x/7 + .....
Let tan x = z, we have
arctan ( z ) = z - z^3/3 + z^5/5 - z^7/7
We thank www.wikipedia.org for publishing this on their site.
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