Trig Identity: Prove cotA + tanA = 2cot2A

[Howard Jay]

I'm really stumped on this identity. Please prove from either or both sides. Anything would help.

Thank You

cotA + tanA = 2cot2A

 

[tkhunny]

Did you switch to sine and cosine and see if anything happens? That's always a good last resort.

In this case, however, it's not all that tricky, since it isn't an identity. Go ahead, try \(\displaystyle A = \frac{\pi}{6}\).

 

[soroban]

Hello, Howard Jay!

\(\displaystyle \cot A \,+\, \tan A \:= \:2\cdot\cot2A\)

The right side is: \(\displaystyle \L\:2\cdot\cot(2A) \;=\;\frac{2}{\tan(2A)} \;=\;\frac{2(1\,-\,\tan^2A)}{2\cdot\tan A} \;=\;\frac{1\,-\,\tan^2A}{\tan A}\)

. . . . . \(\displaystyle \L=\;\frac{1}{\tan A}\,+\,\tan A \;=\;\cot A\,+\,\tan A\)

 

[o_O]

Actually, there's a minor mistake there soroban:
\(\displaystyle \frac{1 - tan^{2}A}{tanA}\)

\(\displaystyle = \frac{1}{tanA} - tanA\)

\(\displaystyle = cotA - tanA\)

which is different from the original "equation". As tkhunny says, it isn't an identity.