Answer: see proof below
Step-by-step explanation:
Note: The given equation cannot be proven. Instead I will prove:
[tex]\cot(2\alpha)=\dfrac{\cot^2\alpha-1}{2\cot \alpha}[/tex]
Use the Reciprocal Identity: cot A = (cos A)/(sin A)
Use the following Double Angle Identities:
cos (2A) = cos² A - sin² A
sin (2A) = 2 sin A · cos A
Proof LHS → RHS:
LHS: cot (2A)
[tex]\text{Reciprocal:}\qquad \qquad \dfrac{\cos (2\alpha)}{\sin (2\alpha)}[/tex]
[tex]\text{Double Angle:}\qquad \quad \dfrac{\cos^2 \alpha-\sin^2 \alpha}{2\sin \alpha \cdot \cos \alpha}[/tex]
[tex]\text{Manipulate:}\qquad \quad \dfrac{\cos^2 \alpha-\sin^2 \alpha}{2\sin \alpha \cdot \cos \alpha}\bigg(\dfrac{\frac{1}{\sin^2 \alpha}}{\frac{1}{\sin^2 \alpha}}\bigg)[/tex]
[tex]=\dfrac{\frac{\cos^2 \alpha}{\sin^2 \alpha}-\frac{\sin^2 \alpha}{\sin^2 \alpha}}{\frac{2\sin \alpa \cdot \cos \alpha}{\sin \alpha \cdot \sin \alpha}}[/tex]
[tex]=\dfrac{\cot^2 \alpha-1}{2\cot \alpha}[/tex]
LHS = RHS [tex]\checkmark[/tex]
