SOLUTION: 2. For all θ where sin θ - cos θ ≠ 0, ((sin^2)(θ) - (cos^2)(θ) / (sinθ - cosθ) is equivalent to...
a.sinθ - cosθ
b.sin_
Algebra.Com
Question 696278: 2. For all θ where sin θ - cos θ ≠ 0, ((sin^2)(θ) - (cos^2)(θ) / (sinθ - cosθ) is equivalent to...
a.sinθ - cosθ
b.sinθ + cosθ
c.tanθ
d.-1
e.1
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
(a\ +\ b)}{a\ -\ b}\ =\ (a\ +\ b)\ \forall\ a,\,b\ \in\ \mathbb{R}, a\ \neq\ b)
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
RELATED QUESTIONS
2 sinθ cosθ =... (answered by Alan3354)
The expression sin θ (cot θ - csc θ) is equivalent to?
A) 2 Cos θ
(answered by stanbon,lynnlo)
The expression sin θ(cotθ - cscθ) is equivalent to
1) cotθ -... (answered by stanbon)
prove each identity
tanθ -1 = sin^2θ-cos^2θ/... (answered by MathLover1)
cosθ/sinθ=1/tanθ (answered by Aswathy)
Prove that:
2cos^3 θ-cos θ/sinθ cos^2 θ-sin^3 θ= cot... (answered by solver91311)
Proof:
2-(sin θ -cos θ )^2= (sinθ+cosθ)^2... (answered by MathLover1)
(sin^2θ -tanθ)/(cos^2θ-cotθ) =... (answered by Edwin McCravy)
Prove that... (answered by Jc0110)
