# Trig Identities : Help 3 Questions.

1. Dec 12, 2004

### xLaser

I can't get these 3 questions, can someone help me?

1. cotB [ (tanA + TanB) / (cotA+cotB) ] = tan A

2. (sin^2A + 2cosA - 1) / (2 + cosA - cos^2A) = 1 / (1+ secA)

3. cos^3A + sin^3A = (cosA+SinA)(1-SinAcosA)

please help out on these, thx in advance. U can write the / sign as fractions cuz i can't do it on the computer here.

2. Dec 12, 2004

### Hurkyl

Staff Emeritus
For proving trig identities, it's generally useful to convert everything into sin and cos, cross-multiply all fractions, and multiply out all factorizations.

3. Dec 12, 2004

### Diane_

It's also helpful to know the double-angle formulas and the half-angle formulas.

4. Dec 12, 2004

### xLaser

no need to use double-angle formulas for those... i just can't figure them out. can someone actually do one?

5. Dec 12, 2004

### kreil

here is the first one:
$$1. tan(A)=cot(B)\frac{tan(A) + tan(B)}{cot(A)+cot(B)}$$

Express in terms of sine and cosine:
$$=(\frac{\cos{B}}{\sin{B}})\frac{\frac{\sin{A}}{\cos{A}}+\frac{\sin{B}}{\cos{B}}}{\frac{cosA}{sinA}+\frac{cosB}{sinB}}$$

Get common denominators and add the top/bottom to form 1 complex fraction:
$$=(\frac{\cos{B}}{\sin{B}})\frac{\frac{sinAcosB+sinBcosA}{cosAcosB}}{\frac{sinBcosA+sinAcosB}{sinAsinB}}$$

Simplify:
$$=(\frac{\cos{B}}{\sin{B}})\frac{\frac{sin(A+B)}{cosAcosB}}{\frac{sin(A+B)}{sinAsinB}}$$

$$=\frac{\frac{sin(A+B)}{cosA}}{\frac{sin(A+B)}{sinA}}=\frac{sinA}{cosA}=tanA$$

Identities used in solution:

$$cotA=\frac{cosA}{sinA}$$

$$tanA=\frac{sinA}{cosA}$$

$$sin(A+B)=sinAcosB+cosAsinB$$

Good luck with the others, I hope this helps you!