Question

please help me.... The question no.b and would like to request you all just give me correct answer. ​

Answer 1

Cos2A = Cos2A
I hope this helps!

Answer 2

Answer: see proof below

Explanation:

You will need the following identities to prove this:

`\tan\ (\alpha-\beta)=(\tan \alpha-\tan \beta)/(1+\tan \alpha\cdot \tan \beta)`

`\cos\ 2\alpha=\cos^2 \alpha-\sin^2\alpha`

LHS → RHS

`(2\tan\ (45^o-A))/(1+\tan^2\ (45^o-A))\\\\\\=(2\bigg((\tan\ 45^o-\tan\ A)/(1+\tan\ 45^o\cdot \tan\ A)\bigg))/(1+\bigg((\tan\ 45^o-\tan\ A)/(1+\tan\ 45^o\cdot \tan\ A)\bigg)^2)\\\\\\=(2\bigg((1-\tan\ A)/(1+\tan\ A)\bigg))/(1+\bigg((1-\tan\ A)/(1+\tan\ A)\bigg)^2)\\\\\\=(2\bigg((1-\tan A)/(1+\tan A)\bigg))/(1+\bigg((1-2\tan\A+\tan^2 A)/(1+2\tan A+\tan^2A)\bigg))\\`

`=(2\bigg((1-\tan A)/(1+\tan A)\bigg))/(((1+2\tan A+\tan^2A)+(1-2\tan A+\tan^2 A))/(1+2\tan A+\tan^2A))\\\\\\=(2\bigg((1-\tan A)/(1+\tan A)\bigg))/((2+2\tan^2A)/(1+2\tan A+\tan^2A))\\\\\\=(2\bigg((1-\tan A)/(1+\tan A)\bigg))/(2\bigg((1+\tan^2A)/((1+\tan A)^2)\bigg))\\\\\\=(\bigg((1-\tan A)/(1+\tan A)\bigg))/(\bigg((1+\tan^2A)/((1+\tan A)^2)\bigg))`

`=(1-\tan A)/(1+\tan A)}* ((1+\tan A)^2)/(1+\tan^2A)\\\\\\=(1-\tan^2 A)/(1+\tan^2 A)\\\\\\=(1-(\sin^2 A)/(\cos^2 A))/(1+(\sin^2 A)/(\cos^2 A))\\\\\\=(\bigg((\cos^2 A-\sin^2 A)/(\cos^2 A)\bigg))/(\bigg((\cos^2 A+\sin^2 A)/(\cos^2 A)\bigg))\\\\\\=(\cos^2 A-\sin^2 A)/(\cos^2 A+\sin^2 A)\\\\\\=(\cos^2 A-\sin^2 A)/(1)\\\\\\=\cos^2 A-\sin^2 A\\\\\\=\cos 2A`

cos 2A = cos 2A
`\checkmark`