Question
If $\tan\text{A}=\frac{1}{7}$ and $\tan\text{B}=\frac{1}{3},$ show that $\cos2\text{A}=\sin4\text{B}$
✓
Answer
We have, $\tan\text{A}=\frac{1}{7}\&\tan\text{B}=\frac{1}{3}$ $\therefore\cos2\text{A}=\frac{1-\tan^2\text{A}}{1+\tan^2\text{A}}=\frac{1-\Big(\frac{1}{7}\Big)^2}{1+\Big(\frac{1}{7}\Big)^2}=\frac{48}{\frac{49}{\frac{50}{49}}}$ $=\frac{48}{50}=\frac{24}{25}\ ...{\text{(A)}}$ Also, $\sin4\text{B}=\sin2.2.\text{B}$ $=2\sin2\text{B}.\cos2\text{B}$ $=2.\Big(\frac{2\tan\text{B}}{1+\tan^2\text{B}}\Big).\Big(\frac{1-\tan^2\text{B}}{1+\tan^2\text{B}}\Big)$ $=4.\Bigg(\frac{\frac{1}{3}}{1+\frac{1}{9}}\Bigg).\Bigg(\frac{1-\frac{1}{9}}{1+\frac{1}{9}}\Bigg)$ $=\frac{4.\frac{1}{3}.\frac{8}{9}}{\frac{10}{9}\times\frac{10}{9}}$ $=\frac{32\times3}{100}$ $=\frac{8\times3}{25}=\frac{24}{25}\ ...(\text{B})$ From(A) & (B) $\cos2\text{A}=\sin4\text{B}$
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