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

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]

Jemma has 24 balls. Out of the 24 balls, 12 are yellow, 4 are pink, and the rest are red. What is the ratio of the number of red balls to the number of balls that are either yellow or pink? arrowRight

Please help! Dad will murder me if I don't have this done by the time he gets home! 100 points for whoever solves this!