Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**jcd29****Member**- Registered: 2008-01-24
- Posts: 1

there's a few problems that I can't solve..maybe you guys can help me:

1) [sin(x-y)] / [cos(x+y)] = [tan(x)-tan(y)] / [1-tan(x)tan(y)]

2) [cot^2(u)-1] / [cot^2(u)+1] = [csc^2(u)-sec^2(u)] / [csc^2(u)+sec^2(u)]

3) [sec^2(u)-tan^2(u)+tan(u)] / sec(u) = sin(u)+cos(u)

It's really urgent..my brother has a test tomorrow and I'm trying to help him, so knowing how to solve this would help a lot. Thanks.

Offline

**saudi_boy****Member**- Registered: 2006-11-01
- Posts: 41

1. [sin(x-y)] / [cos(x+y)] = [tan(x)-tan(y)] / [1-tan(x)tan(y)]

Left side

now divide {numerator}{denominator} by cos(y)cos(x)

Offline

**Monox D. I-Fly****Member**- From: Indonesia
- Registered: 2015-12-02
- Posts: 1,160

jcd29 wrote:

3) [sec^2(u)-tan^2(u)+tan(u)] / sec(u) = sin(u)+cos(u)

3) [sec^2(u)-tan^2(u)+tan(u)] / sec(u)

= [1 + tan(u)] / sec(u)

= cos(u) + sin(u)

= sin(u) + cos(u)

Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away. May his adventurous soul rest in peace at heaven.

Offline

Pages: **1**