Prove that \(\frac{\cos4x+\cos3x+\cos2x}{\sin4x+\sin3x+\sin2x}=\cot3x\)

Prove that:  \(\frac { \sin { \left( 180-\theta \right) } \cos { \left( 90+\theta \right) } \tan { \left( 270-\theta \right) \cot { \left( 360-\theta \right) } } }{ \sin { \left( 360-\theta \right) \cot { \left( 360+\theta \right) \sin { \left( 270-\theta \right) \csc { \left( -\theta \right) } } } } } =-1\)

Prove that \(\tan { \left( \pi +x \right) } \cot { \left( x-\pi \right) } -\left( \cos { \left( 2\pi -x \right) } \cos { \left( 2\pi +x \right) } \right) =\sin ^{ 2 }{ x } \)

If \(\sin A=\frac13,\sin B=\frac14\), then find the value of sin(A + B) where A and B are acute angles.

If \(\cos(\alpha+\beta)=\frac45\) and \(\sin(\alpha-\beta)=\frac{5}{13}\) where \((\alpha+\beta)\) and \((\alpha-\beta)\) are acute, then find \(\tan2\alpha\)