Step-by-step explanation:
To prove that:
[tex]$\frac{1-\cos 2A}{\sin2 A}=\tan A[/tex]
Let us take left hand side and prove it.
[tex]$LHS = \frac{1-\cos 2 A}{\sin 2 A}[/tex]
Using the trigonometric identity: [tex]$\cos (2 x)=1-2 \sin ^{2}(x)[/tex]
[tex]$= \frac{1-(1-2\sin^2A)}{\sin 2 A}[/tex]
Using the trigonometric identity: [tex]\sin (2 x)=2 \cos (x) \sin (x)[/tex]
[tex]$= \frac{1-1+2\sin^2A}{2 \sin A \cos A }[/tex]
[tex]$= \frac{2\sin^2A}{2 \sin A \cos A }[/tex]
[tex]$= \frac{2\sin A \sin A }{2 \sin A \cos A }[/tex]
Cancel the common term 2 sin A on both numerator and denominator.
[tex]$=\frac{\sin A}{\cos A}[/tex]
Using the trigonometric identity: [tex]\frac{\sin (x)}{\cos (x)}=\tan (x)[/tex]
= tan A
= RHS
[tex]$\frac{1-\cos 2A}{\sin2 A}=\tan A[/tex]
Hence proved.
