Question

Prove the following $\cos ec2A-cot2A=\tan A$

Answer 1

Proof:

Taking LHS,

$$\begin{gathered} \mathrm{LHS}=\cos ec2A-cot2A \\ =\frac{1}{\sin 2A}-\frac{\cos 2A}{\sin 2A}\left[\because \mathrm{cosec}A=\frac{1}{\sin A}\mathrm{and}\cot A=\frac{\cos A}{\sin A}\right] \\ =\frac{1-\cos 2A}{\sin 2A} \\ =\frac{2{\sin }^{2}A}{2\sin A·\cos A}\left[\because 1-\cos 2A=2{\sin }^{2}A\mathrm{and}\sin 2A=2\sin A·\cos A\right] \\ =\frac{\sin A}{\cos A} \\ =\tan A \\ =\mathrm{RHS} \end{gathered}$$

Hence, proved.