Question

If $A=ta{n}^{-1}\left(\frac{1}{7}\right)$ and $B=ta{n}^{-1}\left(\frac{1}{3}\right)$, then

• A. $cos2A=\frac{24}{25}$
• B. $cos2B=\frac{4}{5}$
• C. $cos2A=sin4B$
• D. $tan2B=\frac{3}{4}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $tan2B=\frac{3}{4}$
B $cos2A=\frac{24}{25}$
C $cos2B=\frac{4}{5}$
D $cos2A=sin4B$

Given $A=ta{n}^{-1}\left(\frac{1}{7}\right)$ and $B=ta{n}^{-1}\left(\frac{1}{3}\right)$

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

$\cos 2A=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$

$=\frac{1-\frac{1}{49}}{1+\frac{1}{49}}$ $=\frac{48}{50}$ $=\frac{24}{25}$

$\sin 2A=\frac{2\tan A}{1+{\tan }^{2}A}=\frac{\frac{2}{7}}{1+\frac{1}{49}}=\frac{7}{25}$

$\cos 2B=\frac{1-{\tan }^{2}B}{1+{\tan }^{2}B}$ $=\frac{1-\frac{1}{9}}{1+\frac{1}{9}}$ $=\frac{8}{10}$ $=\frac{4}{5}$

$\sin 2B=\frac{2\tan B}{1+{\tan }^{2}B}=\frac{\frac{2}{3}}{1+\frac{1}{9}}=\frac{3}{5}$

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

$\tan 2B$ $=\frac{\sin 2B}{\cos 2B}=\frac{3}{4}$