Prove the identity (cosecA + cotA)^2 = (1+cosA)/(1-cosA)

[Monkeyseat]

Question:

Prove the identity (cosecA + cotA)^2 = (1+cosA)/(1-cosA)

Working:

(cosecA + cotA)^2 = cosec^2 A + 2(cosecA * cotA) + cot^2 A
(cosecA + cotA)^2 = 1/(sin^2 A) + 2((1/sinA) * (cosA/sinA)) + (cos^2 A)/(sin^2 A)
(cosecA + cotA)^2 = (1 + 2cosA + cos^2 A)/(sin^2 A)
(cosecA + cotA)^2 = (1 + 2cosA + cos^2 A)/(1 - cos^2 A)

That's as far as I can get, I can't get to (1+cosA)/(1-cosA). Can someone please help?

Sorry if the formatting is a bit confusing.

Thanks.

 

[soroban]

Hello, Monkeyseat!

You were almost there . . .

Prove the identity: .\(\displaystyle (\csc A + \cot A)^2 \:= \:\frac{1+\cos A}{1-\cos A}\)

Working:

\(\displaystyle (\csc A + \cot A)^2 \;= \;\csc^2\!A + 2\csc A\cot A + \cot^2\!A\)

. . \(\displaystyle = \; \frac{1}{\sin^2\!A} + 2\!\cdot\!\frac{1}{\sin A}\!\cdot\!\frac{\cos A}{\sin A} + \frac{\cos^2\!A}{\sin^2\!A}\)

. . \(\displaystyle = \;\frac{1 + 2\cos A + \cos^2\!A}{\sin^2\!A}\)

. . \(\displaystyle =\; \frac{1 + 2\cos A + \cos^2\!A}{1 -\cos^2\!A}\) . . . . Good!

\(\displaystyle \text{Factor: }\;\frac{(1+\cos A)(1+\cos A)}{(1-\cos A)(1 + \cos A)}\quad\text{and reduce: }\;\frac{1 + \cos A}{1 - \cos A}\)

 

[Deleted member 4993]

Question:

Prove the identity (cosecA + cotA)^2 = (1+cosA)/(1-cosA)

Working:

(cosecA + cotA)^2 = cosec^2 A + 2(cosecA * cotA) + cot^2 A
(cosecA + cotA)^2 = 1/(sin^2 A) + 2((1/sinA) * (cosA/sinA)) + (cos^2 A)/(sin^2 A)
(cosecA + cotA)^2 = (1 + 2cosA + cos^2 A)/(sin^2 A)
(cosecA + cotA)^2 = (1 + 2cosA + cos^2 A)/(1 - cos^2 A) = (1 + cosA)^2/[(1+cosA)(1-cosA)] = (1+cosA)/(1-cosA)

That's as far as I can get, I can't get to (1+cosA)/(1-cosA). Can someone please help?

Sorry if the formatting is a bit confusing.

Thanks.

 

[Monkeyseat]

Thanks both of you, I really appreciate you helping.