Answer: see proof below
Step-by-step explanation:
You will need the following identities to prove this:
[tex]\tan\ (\alpha-\beta)=\dfrac{\tan \alpha-\tan \beta}{1+\tan \alpha\cdot \tan \beta}[/tex]
[tex]\cos\ 2\alpha=\cos^2 \alpha-\sin^2\alpha[/tex]
LHS → RHS
[tex]\dfrac{2\tan\ (45^o-A)}{1+\tan^2\ (45^o-A)}\\\\\\=\dfrac{2\bigg(\dfrac{\tan\ 45^o-\tan\ A}{1+\tan\ 45^o\cdot \tan\ A}\bigg)}{1+\bigg(\dfrac{\tan\ 45^o-\tan\ A}{1+\tan\ 45^o\cdot \tan\ A}\bigg)^2}\\\\\\=\dfrac{2\bigg(\dfrac{1-\tan\ A}{1+\tan\ A}\bigg)}{1+\bigg(\dfrac{1-\tan\ A}{1+\tan\ A}\bigg)^2}\\\\\\=\dfrac{2\bigg(\dfrac{1-\tan A}{1+\tan A}\bigg)}{1+\bigg(\dfrac{1-2\tan\A+\tan^2 A}{1+2\tan A+\tan^2A}\bigg)}\\[/tex]
[tex]=\dfrac{2\bigg(\dfrac{1-\tan A}{1+\tan A}\bigg)}{\dfrac{(1+2\tan A+\tan^2A)+(1-2\tan A+\tan^2 A)}{1+2\tan A+\tan^2A}}\\\\\\=\dfrac{2\bigg(\dfrac{1-\tan A}{1+\tan A}\bigg)}{\dfrac{2+2\tan^2A}{1+2\tan A+\tan^2A}}\\\\\\=\dfrac{2\bigg(\dfrac{1-\tan A}{1+\tan A}\bigg)}{2\bigg(\dfrac{1+\tan^2A}{(1+\tan A)^2}\bigg)}\\\\\\=\dfrac{\bigg(\dfrac{1-\tan A}{1+\tan A}\bigg)}{\bigg(\dfrac{1+\tan^2A}{(1+\tan A)^2}\bigg)}[/tex]
[tex]=\dfrac{1-\tan A}{1+\tan A}}\times \dfrac{(1+\tan A)^2}{1+\tan^2A}\\\\\\=\dfrac{1-\tan^2 A}{1+\tan^2 A}\\\\\\=\dfrac{1-\dfrac{\sin^2 A}{\cos^2 A}}{1+\dfrac{\sin^2 A}{\cos^2 A}}\\\\\\=\dfrac{\bigg(\dfrac{\cos^2 A-\sin^2 A}{\cos^2 A}\bigg)}{\bigg(\dfrac{\cos^2 A+\sin^2 A}{\cos^2 A}\bigg)}\\\\\\=\dfrac{\cos^2 A-\sin^2 A}{\cos^2 A+\sin^2 A}\\\\\\=\dfrac{\cos^2 A-\sin^2 A}{1}\\\\\\=\cos^2 A-\sin^2 A\\\\\\=\cos 2A[/tex]
cos 2A = cos 2A [tex]\checkmark[/tex]
