# Thread: Trig Identities

1. ## Trig Identities

Establish each identity.

NOTE: t = theta for short

(1) csc^4 (t) - csc^2 (t) = cot^4 (t) + cot^2 (t)

(2) csc (t) - cot (t) = [sin(t)]/[1 + cot(t)]

2. For #1, just use the identity $\displaystyle csc^{2}(t)=1+cot^{2}(t)$

Sub into the left side:

$\displaystyle (1+cot^{2}(t))^{2}-(1+cot^{2}(t))$

Now, expand that out and I bet it transforms into the right side.

3. Originally Posted by magentarita
Establish each identity.

NOTE: t = theta for short

(1) csc^4 (t) - csc^2 (t) = cot^4 (t) + cot^2 (t)

(2) csc (t) - cot (t) = [sin(t)]/[1 + cot(t)]
In the trig identity
sin^2(t) +cos^2(t) = 1 -----------(i)
divide both sides by sin^2(t),
1 +cot^2(t) = csc^2(t) -----------(ii)

The first problem:
csc^4 (t) - csc^2 (t) = cot^4 (t) + cot^2 (t)
We play on the LHS to see if it becomes equivalent to the RHS.
LHS = [1 +cot^2(t)]^2 -[1 +cot^2(t)]
= [1 +2cot^2(t) +cot^4(t)] -[1 +cot^2(t)]
= 1 +2cot^2(t) +cot^4(t) -1 -cot^2(t)
= cot^4(t) +cot^2(t)
= RHS
Proven.

------------------------------
The second problem:

csc (t) - cot (t) = [sin(t)] /[1 + cot(t)]

From the derived Eq.(ii) above,
1 = csc^2(t) -cot^2(t) ------------------(iia)

Using a^2 -b^2 = (a+b)(a-b),
We multiply, and divide at the same time, the LHS by [csc(t) +cot(t)],
LHS = csc(t) -cot(t)
= [csc^2(t) -cot^2(t)] / [csc(t) +cot(t)]
= [1] / [{1/sin(t)} +{cos(t)/sin(t)}]
= 1 / [{1 +cos(t)} /sin(t)]
= 1 * [sin(t) /{1 +cos(t)}
= sin(t) / [1 +cos(t)]

Umm, I think there a typo in your posted problem/identity #2. It should be cos(t)....not the posted cot(t)....in the denominator of the RHS.

4. ## great...

Great replies.

Thanks