##### Ads

related to: LHS

**LHS**= (sin^2(x) + 2sinx + 1) / cos^2(x) Now, we know that a^2 + 2ab + b^2 = (a + b)^2 Using the above identity...the identity: a^2 - b^2 = (a - b)(a + b), here: a = 1 and b = sinx] Now substituting (i) and (ii) in**LHS**, we have:**LHS**= (sin^2(x) + 2sinx + 1) / cos^2(x) = (1 + sinx)^2 / [(1 - sinx)(1 + sinx)] = (1 + sinx...3 Answers · Science & Mathematics · 17/02/2008

Take RHS tanx/secx+1=(sinx/cosx)/ (1/cosx+1).............. convert tanx and secx in terms of sinx and cosx =sinx/(1+cosx) =sinx(1-cosx)/(1-cosx^2) =(1-cosx)/sinx.......................................cancel sinx =(secx-1)/tnax...

1 Answers · Science & Mathematics · 28/11/2009

The right hand side is a geometric infinite series that only converges for |x|<1. Outside of this interval, the right hand side diverges. For x=2, the right hand side goes to infinity. The equation you have is only true for |x...

2 Answers · Science & Mathematics · 21/08/2008

{(1 - 2sin^2x) / (cosx + sinx)} = cosx - sinx Consider the left hand side: {(1 - 2sin^2x) / (cosx + sinx)} = {(1 - sin^2x - sin^2x) / (cosx + sinx)} = {(cos^2x - sin^2x) / cosx + sinx)} = {(cosx + sinx)(cosx - sinx) / (cosx + sinx)} = {(cosx - sinx)} = cosx - sinx = right hand side of the equation hence proved

1 Answers · Science & Mathematics · 04/02/2014

1.

**LHS**: taking first component: cosA / (1- ...) and so we are left with sinA + cosA,**LHS**= RHS as required. 2.**LHS**tan^2(θ) + cot...2 Answers · Science & Mathematics · 09/07/2012

cause to get the answer you should multiply (x+2) thrice to itself, not just put an exponent of 3 in each term. (x+2)^3 = (x+2)(x+2)(x+2) = (x^2 + 4x + 4)(x + 2) = (x^3 + 2x^2 + 4x^2 + 8x + 4x + 8) = x^3 + 6x^2 + 12x + 8

3 Answers · Science & Mathematics · 17/09/2011

You have a hint. The integral has something to do with a trigonometric function. So, let's work from there Let x = a * sin(t), then dx would equal a * cos(t) * dt dx / sqrt(a^2 - x^2) => a * cos(t) * dt / sqrt(a^2 - a^2 * sin(t)^2) => a * cos(t) * dt / (sqrt(a^2 * (1 - sin(t)^2)) => a * cos(t) * dt...

2 Answers · Science & Mathematics · 13/12/2011

(27^x + 3) cannot "simplify to" 69(3^(2x) + 1). It is true that 27^x+3=3{3^(2x)+1}

1 Answers · Science & Mathematics · 24/06/2011

Here's what I got from this: Your first trig identity, which says [A] sin^2(x) + cos^2(x) = 1, can be rewritten as [B] cos^2(x) = 1 - sin^2(x), so my first step would be: [1 - sin^2(x)] / [1 + cot^2(x)] [cos^2(x)] / [1 + cot^2(x...

2 Answers · Science & Mathematics · 07/05/2011

a+b/ab

1 Answers · Science & Mathematics · 17/09/2011

##### Ads

related to: LHS