Question

Let $\cos \left( \alpha +\beta \right)=\dfrac{4}{5}$ and let $\sin \left( \alpha -\beta \right)=\dfrac{5}{13}$ , where $0\le \alpha ,\beta \le \dfrac{\pi }{4}$ then $\tan \left( 2\alpha \right)=$
$\begin{align}
  & a)\dfrac{56}{33} \\
 & b)\dfrac{19}{12} \\
 & c)\dfrac{20}{7} \\
 & d)\dfrac{25}{16} \\
\end{align}$

Answer 1

Hint: Now we are given with $\cos \left( \alpha +\beta \right)$ and $\sin \left( \alpha -\beta \right)$.
Now we know that ${{\sec }^{2}}x=\dfrac{1}{{{\cos }^{2}}x}$ , ${{\tan }^{2}}x+1={{\sec }^{2}}x$ and ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ . Using these identities we can find the value of $\tan \left( \alpha +\beta \right)$ and $\tan \left( \alpha -\beta \right)$ . Now we have $\tan \left( 2\alpha \right)=\tan \left( \alpha +\beta +\left( \alpha -\beta \right) \right)$ and $\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$ . hence we can get the value of $\tan 2\alpha $

Complete step by step answer:
Now first consider $\cos \left( \alpha +\beta \right)=\dfrac{4}{5}$
Squaring the equation on both sides we get
${{\cos }^{2}}\left( \alpha +\beta \right)=\dfrac{16}{25}$
Taking inverse on both sides we get
$\dfrac{1}{{{\cos }^{2}}\left( \alpha +\beta \right)}=\dfrac{25}{16}$
Now we know that ${{\sec }^{2}}x=\dfrac{1}{{{\cos }^{2}}x}$
Hence we get
${{\sec }^{2}}\left( \alpha +\beta \right)=\dfrac{25}{16}$
Now we have identity ${{\tan }^{2}}x+1={{\sec }^{2}}x$ using this we get.
$\begin{align}
  & 1+{{\tan }^{2}}\left( \alpha +\beta \right)=\dfrac{25}{16} \\
 & \Rightarrow {{\tan }^{2}}\left( \alpha +\beta \right)=\dfrac{25}{16}-1 \\
 & \Rightarrow {{\tan }^{2}}\left( \alpha +\beta \right)=\dfrac{25-16}{16}=\dfrac{9}{16} \\
 & \Rightarrow \tan \left( \alpha +\beta \right)=\pm \dfrac{3}{4} \\
\end{align}$
Now since we have $0\le \alpha ,\beta \le \dfrac{\pi }{4}$ we can say $0\le \alpha +\beta \le \dfrac{\pi }{2}$ and tan is positive in first quadrant.
Hence we get $\tan \left( \alpha +\beta \right)=\dfrac{3}{4}...............(1)$
Now consider $\sin \left( \alpha -\beta \right)=\dfrac{5}{13}$
Now squaring on both sides we get ${{\sin }^{2}}\left( \alpha -\beta \right)=\dfrac{25}{169}$
Now we know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$
Hence we get \[{{\cos }^{2}}\left( \alpha -\beta \right)+\dfrac{25}{69}=1\]
Hence we get
\[\begin{align}
  & {{\cos }^{2}}\left( \alpha -\beta \right)=1-\dfrac{25}{169} \\
 & \Rightarrow {{\cos }^{2}}\left( \alpha -\beta \right)=\dfrac{169-25}{169}=\dfrac{144}{169} \\
\end{align}\]
Now if we take inverse on both sides we get.
$\dfrac{1}{{{\cos }^{2}}\left( \alpha -\beta \right)}=\dfrac{169}{144}$
Now we have ${{\sec }^{2}}x=\dfrac{1}{{{\cos }^{2}}x}$ using this identity we get
${{\sec }^{2}}\left( \alpha -\beta \right)=\dfrac{169}{144}$
Now we have the identity we get ${{\tan }^{2}}x+1={{\sec }^{2}}x$ . hence we get
$\begin{align}
  & 1+{{\tan }^{2}}\left( \alpha -\beta \right)=\dfrac{169}{144} \\
 & \Rightarrow {{\tan }^{2}}\left( \alpha -\beta \right)=\dfrac{169}{144}-1 \\
 & \Rightarrow {{\tan }^{2}}\left( \alpha -\beta \right)=\dfrac{169-144}{144}=\dfrac{25}{144} \\
 & \Rightarrow \tan \left( \alpha -\beta \right)=\pm \dfrac{5}{12} \\
\end{align}$
Now we have $0\le \alpha ,\beta \le \dfrac{\pi }{4}$
Hence $\dfrac{-\pi }{4}\le \alpha -\beta \le \dfrac{\pi }{4}$
Now the angle $\alpha -\beta $ lies in the fourth of the first quadrant.
Now since $\sin \left( \alpha -\beta \right)$ is positive we have $\alpha -\beta $ is in first quadrant
And we know that the first quadrant tan is positive.
Hence we have
$\tan \left( \alpha -\beta \right)=\dfrac{5}{12}.......................\left( 2 \right)$
Now consider $\tan \left( \alpha +\beta +\left( \alpha -\beta \right) \right)$
Now we know that $\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$
Hence we get
\[\tan \left( \alpha +\beta +\left( \alpha -\beta \right) \right)=\dfrac{\tan \left( \alpha +\beta \right)+\tan \left( \alpha -\beta \right)}{1-\tan \left( \alpha +\beta \right)\tan \left( \alpha -\beta \right)}\]
Now from equation (1) and equation (2) we get
\[\tan \left( \alpha +\beta +\left( \alpha -\beta \right) \right)=\dfrac{\dfrac{3}{4}+\dfrac{5}{12}}{1-\left( \dfrac{3}{4} \right)\left( \dfrac{5}{12} \right)}\]
Now taking LCM we get
\[\begin{align}
  & \tan \left( \alpha +\beta +\alpha -\beta \right)=\dfrac{\dfrac{3\times 3}{4\times 3}+\dfrac{5}{12}}{1-\left( \dfrac{15}{48} \right)} \\
 & \Rightarrow \tan \left( \alpha +\beta +\alpha -\beta \right)=\dfrac{\dfrac{9+5}{12}}{\left( \dfrac{48-15}{48} \right)} \\
 & \Rightarrow \tan \left( 2\alpha \right)=\dfrac{\dfrac{14}{1}}{\dfrac{33}{4}}=\dfrac{14\times 4}{33}=\dfrac{56}{33} \\
\end{align}\]
Hence the value of $\tan \left( 2\alpha \right)=\dfrac{56}{33}$

So, the correct answer is “Option A”.

Note: Now we in such sums we can skip evaluating each trigonometric ratio with the help of identities. We know that tan is the ratio of opposite side and adjacent side. Now we can find the length if hypotenuse using Pythagoras theorem. Once we have the length of all three sides we can easily find other trigonometric ratios.