- #1

Cosmic-Kat

- 8

- 0

I have re-post this forum as I should have paid closer attention to rules. I apologized for that.

Tan=sin/cos

Sec=1/cos

Cot= cos/sin

Csc= 1/Sin

(2Cos/(1+Sin)(Sin-1))^-1=

(2Cos(Sin-1)/(1+Sin))^-1=

(-2Cos/1)^-1=

1/2cos= 2sec

(?)

(b) 2Sec

The third question:

(Cos/Sin)/ (1/Sin) -(Sin/1) =

(Cos/Sin)(Sin/1)/ (1/Sin)=

(Cos*Sin/Sin)/(1/Sin)=

Cos/(1/Sin) =

Sin/Cos = Tan

(c) Tan

## Homework Statement

*1) The expression tan^3 θ + sinθ/cosθ is equal to:*

(a) cot θ (b) tan θ sec^2 θ (c) tan θ (d) sin θ tan θ (e) tan θ csc^2 θ 2) Simplify (cos θ/1+ sin θ - cosθ/sinθ-1)^-1

(a) cos θ/2 (b) 2sec θ (c) 2sin θ (d) csc θ/2 3) The expression cot θ/csc θ-sin θ is equal to:

(a) cos θ (b) sec θ (c) tan θ (d) sin θ (e) csc(a) cot θ (b) tan θ sec^2 θ (c) tan θ (d) sin θ tan θ (e) tan θ csc^2 θ 2) Simplify (cos θ/1+ sin θ - cosθ/sinθ-1)^-1

(a) cos θ/2 (b) 2sec θ (c) 2sin θ (d) csc θ/2 3) The expression cot θ/csc θ-sin θ is equal to:

(a) cos θ (b) sec θ (c) tan θ (d) sin θ (e) csc

## Homework Equations

:[/B]*Identifying the trigonometry identities*Tan=sin/cos

Sec=1/cos

Cot= cos/sin

Csc= 1/Sin

## The Attempt at a Solution

:[/B]Here's my attempt on the first question:*Tan^3 = Sin^3/Cos^3;*

Sin^3/Cos^3 + Sin/Cos=

Sin^4/Cos^4 = TanSin^3/Cos^3 + Sin/Cos=

Sin^4/Cos^4 = Tan

*(c) Tan**The second question (I'm very lost on this) :*

(2Cos/(1+Sin)(Sin-1))^-1=

(2Cos(Sin-1)/(1+Sin))^-1=

(-2Cos/1)^-1=

1/2cos= 2sec

(?)

(b) 2Sec

The third question:

(Cos/Sin)/ (1/Sin) -(Sin/1) =

(Cos/Sin)(Sin/1)/ (1/Sin)=

(Cos*Sin/Sin)/(1/Sin)=

Cos/(1/Sin) =

Sin/Cos = Tan

(c) Tan

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