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If tan


If $\tan A=\frac{1}{7}$ and $\tan B=\frac{1}{3}$, show that $\cos 2 A=\sin 4 B$



$\tan A=\frac{1}{7}$ and $\tan B=\frac{1}{3}$

Using the identity $\tan 2 B=\frac{2 \tan B}{1-\tan ^{2} B}$, we get

$\tan 2 B=\frac{2 \times \frac{1}{3}}{1-\frac{1}{9}}=\frac{3}{4}$

Now, using the identities $\cos 2 A=\frac{1-\tan ^{2} A}{1+\tan ^{2} A}$ and $\sin 4 B=\frac{2 \tan 2 B}{1+\tan ^{2} 2 B}$, we get

$\cos 2 A=\frac{1-\left(\frac{1}{7}\right)^{2}}{1+\left(\frac{1}{7}\right)^{2}}$ and $\sin 4 B=\frac{2 \times \frac{3}{4}}{1+\left(\frac{3}{4}\right)^{2}}$

$\Rightarrow \cos 2 A=\frac{48}{50}$ and $\sin 4 B=\frac{2 \times \frac{3}{4} \times 16}{25}$

$\Rightarrow \cos 2 A=\frac{24}{25}$ and $\sin 4 B=\frac{24}{25}$

$\therefore \cos 2 A=\sin 4 B$

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