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1. ### Prove trigonometric identities?

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

2. ### help me verify this integral dx/ sqrt(a^2 -x^2) = arcsin(x/a) + c [so just turn the RHS into the LHS]?

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

3. ### By LHS manipulation ONLY... show that (27^x + 3) simplifies to 69(3^(2x) + 1)?

(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

4. ### Prove [(1-sin^2(x)]/[(1+cot^2(x)] = sin^2(x) + cos^2(x) find LHS and RHS?

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

5. ### 1/a + 1/b = 2/a+b - Pleass explain why they are incorrect and show the correct algebra working from LHS?

a+b/ab

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

6. ### prove these identities?

LHS = sec^2 x - cosec^2 x 1/cos^2 x - 1/sin^2 x = RHS tan^2x - cot^2x...1 - [1/sin^2x - 1] = 1/cos^2x - 1/sin^2x - 1 + 1 = 1/cos^2x - 1/sin^2x so LHS = RHS b) LHS sec x + tan x = 1/cos x + sinx /cos x = (1 + sin...

3 Answers · Science & Mathematics · 20/09/2007

7. ### how to integrate dy/dx=2-3y^4..I try to do like bring to LHS and become dy/(2-3y^4)=dx and solve as usual .my?

dy/(2-3y^4)= use partial fractions then integrate

1 Answers · Science & Mathematics · 26/03/2011

8. ### Prove [tan(x)-cot(x)]/[tan(x)+cot(x) = sin^2(x) cos^2(x) find LHS and RHS?

[tan(x)-cot(x)]/[tan(x)+cot(x)] = sin^2(x)-cos^2(x) Expand to expressions of sine and cosine. [sin(x)/cos(x) - cos(x)/sin(x)] / [sin(x)/cos(x) + cos(x)/sin(x)] = sin^2(x)-cos^2(x) Find the common (denominator) of both the Denominator and the Numerator. For both, it will be sin(x)cos...

1 Answers · Science & Mathematics · 10/05/2011

9. ### Prove [sin^4(x)]-[(cos^4(x))/(sin^2(x))]= 2-csc^2(x) find LHS and RHS?

Hint: the top is a^4 - b^4 = [ a ² + b ² ] [ a ² - b ² ] for appropriate a & b

1 Answers · Science & Mathematics · 08/05/2011

10. ### Prove [tan^2(x)/1+tan^2(x)=sin^2(x) Find LHS and RHS?

From the identity sin^2(x) + cos^2(x) = 1, we have 1 + tan^2(x) = sec^2(x), by dividing by cos^2(x). tan^2(x)/ (1 + tan^2(x) ) = tan^2(x) / sec^2(x) = tan^2(x) * cos^2(x) = (sin^2(x)/cos^2(x)) * cos^2(x) = sin^2(x) Anytime you see tan^2, cot^2, sec^2, or csc^2, recall the identity sin^2(x) + cos^2(x) = 1, and divide by either sin^2(x...

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