# Math Help - Trig Identities - Simplify and Prove

1. ## Trig Identities - Simplify and Prove

I am dying here trying to solve this problem, I'd appreciate if you could help me

Simplify:
g(x) = 1/(4sin^2x)(cos^2x) - (1-tan^2x)^2/4tan^2x

Prove:
(cotB/cscB-1) + (cotB/cscB+1) = 2secB

(tan^3x/1+tan^2x) + (cot^3x/1+cot^2x) = 1-2sin^2xcos^2x/sinxcosx

If you get these your a math god and I will love you for the rest of my life

~Thanks

RedBooster

2. (cotB/cscB-1) + (cotB/cscB+1) = 2secB

LS
[cotB(cscB+1)+cotB(cscB-1)]/[(cscB)^2-1]
(cotBcscB+cotB+cotBscscB-cotB)/(cotB)^2
2cotBcscB/(cotB)^2
2cscB/(cotB)
2(1/sinB)/(cosB/sinB)
2(1/sinB)(sinB/cosB)
2(1/cosB)
2secB = RS

I will solve the other ones later

3. Hello, RedBooster!

The last one is a killer . . .

Prove: . $\frac{\tan^3\!x}{1+\tan^2\!x} + \frac{\cot^3\!x}{1+\cot^2\!x} \;=\;\frac{1-2\sin^2\!x\cos^2\!x}{\sin x\cos x}$

The left side is: . $\frac{\dfrac{\sin^3\!x}{\cos^3\!x}} {1 + \dfrac{\sin^2\!x}{\cos^2\!x}} + \frac{\dfrac{\cos^3\!x}{\sin^3\!x}} {1 + \dfrac{\cos^2\!x}{\sin^2\!x}}$

. . $= \; \frac{{\color{blue}\cos^3\!x}\left(\dfrac{\sin^3\! x}{\cos^3\!x}\right)} {{\color{blue}\cos^3\!x}\left(1 + \dfrac{\sin^2\!x}{\cos^2\!x}\right)} + \frac{{\color{blue}\sin^3\!x}\left(\dfrac{\cos^3\! x}{\sin^3\!x}\right)} {{\color{blue}\sin^3\!x}\left(1 + \dfrac{\cos^2\!x}{\sin^2\!x}\right)}$

. . $= \;\frac{\sin^3\!x}{\cos^3\!x+\sin^2\!x\cos x} + \frac{\cos^3\!x}{\sin^3\!x + \sin x\cos^2\!x}$

. . $= \;\frac{\sin^3\!x}{\cos x\underbrace{(\sin^2\!x+\cos^2\!x)}_{\text{This is 1}}} + \frac{\cos^3\!x}{\sin x\underbrace{(\sin^2\!x+\cos^2\!x)}_{\text{This is 1}}}$

. . $= \;\frac{\sin^3\!x}{\cos x} + \frac{\cos^3\!x}{\sin x}$

. . $=\;{\color{blue}\frac{\sin x}{\sin x}}\cdot\frac{\sin^3\!x}{\cos x} + {\color{blue}\frac{\cos x}{\cos x}}\cdot\frac{\cos^3\!x}{\sin x}$

. . $=\;\frac{\sin^4\!x + \cos^4\!x}{\sin x\cos x}$

. . $=\;\frac{\sin^4\!x \:{\color{red}+ \:2\sin^2\!x\cos^2\!x} + \cos^4\!x \:{\color{red}- \:2\sin^2\!x\cos^2\!x}}{\sin x\cos x}$

. . $= \;\frac{(\overbrace{\sin^2\!x + \cos^2\!x}^{\text{This is 1}})^2 - 2\sin^2\!x\cos^2\!x}{\sin x\cos x}$

. . $= \;\frac{1 - 2\sin^2\!x\cos^2\!x}{\sin x\cos x} \quad\hdots$ . ta-DAA!

4. Wow, thank you soo much! This really has helped me