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camilasanchez
verify identity in terms of sine and cosines
\[\frac{ 1+ \cos^2 3 \theta }{ \sin^2 3 \theta } = 2 \csc^2 3 \theta \]
Because 3θ is everywhere, it plays no special part. No need to break down into smaller angles. Just rewrite 1+cos²3θ as 1+1-sin²3θ=2-sin²3θ, so the left hand side becomes:\[\frac{ 2-\sin^23θ }{ \sin^2 3θ }=\frac{ 2 }{ \sin^23θ }-1\] Hmm... I see a problem here: right hand side is equal to:\[2\csc^23θ=2 \cdot \frac{ 1 }{ \sin^23θ }=\frac{ 2 }{ \sin^23θ }\]So this "identity" seems to be false! We're left with the -1...