Question

If $A = {\tan ^{ - 1}}\left( {\dfrac{1}{7}} \right)$ and $B = {\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right)$ , then
(A) $\cos 2A = \dfrac{{24}}{{25}}$
(B) $\cos 2B = \dfrac{4}{5}$
(C) $\cos 2A = \sin 4B$
(D) $\tan 2B = \dfrac{3}{4}$

Answer 1

Hint:
In this question first we will convert $A$ into $\tan A$ and $B$ into $\tan B$ .Now, we will use the identity of $\cos 2\theta $ and $\sin 2\theta $ in terms of $\tan \theta $ and then by substituting the value of $\tan A$ in the formula we will check each options one by one. Hence, by doing so we will find the correct option.

Formula used:
The formula we are going to use in this question is $\cos 2\theta = \dfrac{{1 - {{\tan }^2} \theta }}{{1 + {{\tan }^2}\theta }}$ and $\sin 2 \theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2} \theta }}$ .

Complete step by step solution:
The given values are $A = {\tan ^{ - 1}}\left( {\dfrac{1}{7}} \right)$ and $B = {\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right)$ . Now, we can write $A = {\tan ^{ - 1}}\left( {\dfrac{1}{7}} \right)$ as $\tan A = \dfrac{1}{7}$ similarly we can write $\tan B = \dfrac{1}{3}$ .
Now use the formula $\cos 2A = \dfrac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}$ and substitute the value of $\tan A = \dfrac{1}{7}$ in the formula. Therefore, we can write.
$
  \cos 2A = \dfrac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} \\
   \Rightarrow \cos 2A = \dfrac{{1 - {{\left( {\dfrac{1}{7}} \right)}^2}}}{{1 + {{\left( {\dfrac{1}{7}} \right)}^2}}} = \dfrac{{1 - \dfrac{1}{{49}}}}{{1 + \dfrac{1}{{49}}}} \\
 $
Simplify, the above equation
$\Rightarrow \dfrac{\dfrac{48}{49}}{\dfrac{50}{49}} \Rightarrow \dfrac{48}{49} \times \dfrac{49}{50} \Rightarrow \dfrac{48}{50} \Rightarrow \dfrac{24}{25}$
Now, substitute the value of $\tan B = \dfrac{1}{3}$ in the formula $\cos 2B = \dfrac{{1 - {{\tan }^2}B}}{{1 + {{\tan }^2}B}}$ . Therefore, we can write
$
  \cos 2B = \dfrac{{1 - {{\tan }^2}B}}{{1 + {{\tan }^2}B}} \\
   \Rightarrow \cos 2B = \dfrac{{1 - {{\left( {\dfrac{1}{3}} \right)}^2}}}{{1 + {{\left( {\dfrac{1}{3}} \right)}^2}}} = \dfrac{{1 - \dfrac{1}{9}}}{{1 + \dfrac{1}{9}}} \\
 $
Simplify the above equation
$ \Rightarrow \dfrac{{{{8} {\left/
 { {}}\right.}
{9}}}}{{{{{10}} {\left/
 { { 9}}\right.}
{}}}} = \dfrac{8}{{10}} = \dfrac{4}{5}$
Now, similarly we will find the value of $\sin 2B$ with the help of this formula $\sin 2B = \dfrac{{2\tan B}}{{1 + {{\tan }^2}B}}$ . Substitute the value of $\tan B = \dfrac{1}{3}$ in $\sin 2B = \dfrac{{2\tan B}}{{1 + {{\tan }^2}B}}$ . Therefore, we can write
$
   \Rightarrow \sin 2B = \dfrac{{2\tan B}}{{1 + {{\tan }^2}B}} \\
   \Rightarrow \sin 2B = \dfrac{{2\left( {\dfrac{1}{3}} \right)}}{{1 + {{\left( {\dfrac{1}{3}} \right)}^2}}} = \dfrac{{{{2} {\left/
 { {}}\right.}
{3}}}}{{1 + \dfrac{1}{9}}} \\
 $
Now, simplify the above equation
$ \Rightarrow \sin 2B = \dfrac{{{{2} {\left/
 { {}}\right.}
{3}}}}{{{{{10}} {\left/
 { {}}\right.}
{9}}}} = \dfrac{3}{5}$
Now, from the above calculation we say that the option (A) and the option (B) are correct. Now, we will check the option (C)
We know that $\sin 2\theta = 2\sin \theta \cos \theta $ . Therefore, we can write $\sin 4B = 2\sin 2B\cos 2B$ . Now, substitute the value of $\sin 2B$ and $\cos 2B$ in $\sin 4B = 2\sin 2B\cos 2B$ . Therefore, we can write:
$
   \Rightarrow \sin 4B = 2\sin 2B\cos 2B \\
   \Rightarrow \sin 4B = 2\left( {\dfrac{3}{5}} \right)\left( {\dfrac{4}{5}} \right) = \dfrac{{24}}{{25}} \\
 $
Now, we can say that $\cos 2A = \sin 4B = \dfrac{{24}}{{25}}$
For option (D) we can write $\tan 2B = \dfrac{{\sin 2B}}{{\cos 2B}} = \dfrac{\dfrac{3}{5}}{\dfrac{4}{5}} \Rightarrow \dfrac{3}{4}$

Therefore, we can say that option (C) and option (D) are correct options. Hence, in this question all the four options are correct.

Note:
In this question one of the important things is the conversion of $A$ into $\tan A$ and $B$ into $\tan B$ because this conversion will allow us to use the identity very effectively. The other important thing is that we have to check every option because in this question multiple answers are correct.