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Verify the Trigonometry Identity. Appreciate the help! :D
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\[2\tan(\cos^-1 (x+1)/2)=(\sqrt{16-4(x+1)^2})/(x+1)\]
cos^-1=arccos
okay so this requires a few trig identities.
\[1+\tan^2(b)=\sec^2(b)=\frac{ 1 }{ \cos^2(b) }\]
where \[b=\cos^{-1}(\frac{x+1}{2})\]
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Now we take the cos of both sides to obtain: \[\cos(b)=\frac{x+1}{2}\]
Plug that into the trig identity above: \[1+\tan^2(b)=\frac{1}{\cos^2(b)}=\frac{1}{(\frac{x+1}{2})^2}\]
Can you see where to go from here?
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