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1. ### Prove (1 - 2 sin^2 x) / (1 + sin 2x) = (1 - sin 2x) / (1 - 2 sin^2 x)?

By multiplying the LHS by [1 - sin(2x)]/[1 - sin(2x)], we see that: LHS = [1 - 2sin^2(x)]/[1 + sin(2x...

1 Answers · Science & Mathematics · 04/05/2011

2. ### Prove by induction ∑ i*2^i = 2 + (n-1) * 2^(n+1) for n >= 1. Index of summation is i = 1 .. n?

WHEN n = 1 LHS = 2 and RHS = 2 so correct n = 2 LHS = 2 + 8 = 10 RHS = 2...

2 Answers · Science & Mathematics · 17/04/2008

3. ### prove each identity.?

We have: a) LHS = cosθ + sinθtanθ = cosθ + sinθ(sinθ/cosθ) . . . . .tanθ = sinθ/cosθ = cos...tanθ. . . . . . . . . . . . . . . . . . . . . . . . . Tangent Identity = LHS I hope this helps!

4 Answers · Science & Mathematics · 04/01/2010

4. ### Help with sum/difference trig question?

LHS= cos(3pi+x)= cos(2pi+pi+x)= cos(pi+x)= -cosx=..........(formula) RHS

3 Answers · Science & Mathematics · 16/12/2009

Method - 1 ........................................................................................... LHS = ( 1 + tan x ) / ( 1 + cot x ) ....... = ( 1 + tan x ) / [ 1 + ( 1 / tan x...

1 Answers · Science & Mathematics · 12/01/2010

LHS = a cos(pi/4) = 4root2 ---. a = 8

1 Answers · Science & Mathematics · 15/12/2010

7. ### ELIMINATION MATH HELPPP?

Putting all on the LHS and re-arranging we have 5x^2-4x+5y^2+20y+19=0. Putting the 5 in...

2 Answers · Science & Mathematics · 25/11/2010

8. ### prove 1) [tan(x-y) +tan y] / [1-tan(x-y) tan y] = tan x?

piece of cake. LHS = [ tan ( x - y ) + tany ] / [ 1 - tan ( x - y ) tan y ] LHS = tan ( x - y + y ) LHS = tan x LHS = RHS QED you do the other one.

1 Answers · Science & Mathematics · 28/06/2013

9. ### grade 11 trig identity problem...?

Taking the LHS and converting it to the RHS LHS = sin^2(x) - sin^4(x...

2 Answers · Science & Mathematics · 17/05/2010

10. ### Math Identity Question?

Does the LHS equal 2tanx / (1+tan^2 x) ? = 2 tan x / sec^2 x = (2 sin x / cos x ) / ( 1/ cos^2 x) = 2 sin x /cos x * cos^2 x = 2 sin x cos x = sin 2x = RHS NB Check your question

1 Answers · Science & Mathematics · 05/12/2012