cbse class X (class 10) exam result for chennai region has been announced .
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index of math problems
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Monday, May 21, 2012
Saturday, February 11, 2012
if A+B+C = 180° , show that cot A cotB + cotBcotC +cotC cotA=1
if A+B+C = 180° ,
show that
tanA +tanB + tanC = tanAtanBtanC and
cot A cotB + cotBcotC +cotC cotA=1
given A+B+C = 180°
implies A + B = 180°- C
implies tan(A + B) = tan(180°- C)
using trigonometry formulae
tan(A + B) = -tan(C)
therefore
[tanA +tanB] / [1-tanAtanB ] = -tanC
cross multiplying
tanA +tanB = -tanC[1-tanAtanB ]
tanA +tanB = -tanC +tanAtanBtanC
or
tanA +tanB + tanC = tanAtanBtanC
divide each term with tanAtanBtanC we get
cot A cotB + cotBcotC +cotC cotA=1
some other problems
If A+B+C=π,
prove that: sinA+sinB+sinC = cot(A/2).cot(B/2) [sinA+sinB-sinC]
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show that
tanA +tanB + tanC = tanAtanBtanC and
cot A cotB + cotBcotC +cotC cotA=1
given A+B+C = 180°
implies A + B = 180°- C
implies tan(A + B) = tan(180°- C)
using trigonometry formulae
tan(A + B) = -tan(C)
therefore
[tanA +tanB] / [1-tanAtanB ] = -tanC
cross multiplying
tanA +tanB = -tanC[1-tanAtanB ]
tanA +tanB = -tanC +tanAtanBtanC
or
tanA +tanB + tanC = tanAtanBtanC
divide each term with tanAtanBtanC we get
cot A cotB + cotBcotC +cotC cotA=1
some other problems
If A+B+C=π,
prove that: sinA+sinB+sinC = cot(A/2).cot(B/2) [sinA+sinB-sinC]
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index of math problems
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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work
Saturday, February 4, 2012
some useful symbols
≡ ~
± ∓
∤ ◅
∈ ∉
⊆ ⊂ ∪ ∩ ⊥ ô
≈
≅
≠
≤
≥
·
∫ Σ → ∞
Π Δ Φ Ψ Ω Γ ∮ ∇∂ √ ∅ °
α β γ δ ε ζ η θ
ι κ λ μ ν ξ ο π
ρ σ τ υ φ χ ψ ω x²
some interesting symbols for posting into forums.
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index of math problems
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± ∓
∤ ◅
∈ ∉
⊆ ⊂ ∪ ∩ ⊥ ô
≈
≅
≠
≤
≥
·
∫ Σ → ∞
Π Δ Φ Ψ Ω Γ ∮ ∇∂ √ ∅ °
α β γ δ ε ζ η θ
ι κ λ μ ν ξ ο π
ρ σ τ υ φ χ ψ ω x²
some interesting symbols for posting into forums.
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index of math problems
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Friday, January 20, 2012
confidence interval
confidence interval for population proportion (P) is
( p - (z)sqrt(pq/n) , p + (z)sqrt(pq/n) )
where z is the table value corresponding to the required confidence level, (two tailed)
p the sample proportion , q = 1 - p
n = sample size
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index of math problems
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( p - (z)sqrt(pq/n) , p + (z)sqrt(pq/n) )
where z is the table value corresponding to the required confidence level, (two tailed)
p the sample proportion , q = 1 - p
n = sample size
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index of math problems
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Thursday, January 19, 2012
inverse and direct variation
u is said to be directly proportional to v if u = c v for some constant c
u is said to be inversely proportional to v if u = [k / v] for some constant k
In direct variation, u increases as v increases and vice versa.
In inverse variation, u decreases as v increases and vice versa.
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u is said to be inversely proportional to v if u = [k / v] for some constant k
In direct variation, u increases as v increases and vice versa.
In inverse variation, u decreases as v increases and vice versa.
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index of math problems
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Monday, January 2, 2012
problem of splitting into real and imaginary parts
if u+iv = (x +iy) ^3 , show that (u/x) + (v/y) = 4( x² -y² )
first expand (x +iy) ^3 using binomial expansion or identity and then split into real and imaginary parts then manipulate

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first expand (x +iy) ^3 using binomial expansion or identity and then split into real and imaginary parts then manipulate

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Saturday, December 31, 2011
integral of (cos x)^3
integral of (cos x)^3
the (cosx)^3 is split into (cos²x )cosx
the cos²x is written as 1 - sin²x (using trigonometry formulae)
then the substitution t = sinx , dt=cosx dx is used
more explanation is given below

click here for integration formulae
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the (cosx)^3 is split into (cos²x )cosx
the cos²x is written as 1 - sin²x (using trigonometry formulae)
then the substitution t = sinx , dt=cosx dx is used
more explanation is given below

click here for integration formulae
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please leave your comments below
------------------------------------------------------------
index of math problems
disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work
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