Question

If $\sin \left(A-B\right)=\frac{1}{\sqrt{10}}$, $\cos \left(A+B\right)=\frac{2}{\sqrt{29}}$ find $\tan \left(2A\right)$ where $A$ and $B$ lie between $0$ to $\frac{\pi }{4}$.

Answer 1

Step 1: Use trigonometric formulas.

Given: $\sin \left(A-B\right)=\frac{1}{\sqrt{10}}$ and $\cos \left(A+B\right)=\frac{2}{\sqrt{29}}$.

We know that, $\sin \theta =\sqrt{1-{\cos }^2\theta }$.

So,

$$\begin{gathered} \sin \left(A+B\right)=\sqrt{1-{\cos }^2\left(A+B\right)} \\ \Rightarrow \sin \left(A+B\right)=\sqrt{1-{\left(\frac{2}{\sqrt{29}}\right)}^2} \\ \Rightarrow \sin \left(A+B\right)=\frac{5}{\sqrt{29}} \end{gathered}$$

We know that, $\cos \theta =\sqrt{1-{\sin }^2\theta }$.

So,

$$\begin{gathered} \cos \left(A-B\right)=\sqrt{1-{\sin }^2\left(A-B\right)} \\ \Rightarrow \cos \left(A-B\right)=\sqrt{1-{\left(\frac{1}{\sqrt{10}}\right)}^2} \\ \Rightarrow \cos \left(A-B\right)=\frac{3}{\sqrt{10}} \end{gathered}$$

Step 2: Find the value of the tangent of the sum and the difference of $A$ and $B$.

Since, $\cos \left(A+B\right)=\frac{2}{\sqrt{29}}$ and $\sin \left(A+B\right)=\frac{5}{\sqrt{29}}$.

So,

$$\begin{gathered} \tan \left(A+B\right)=\frac{\sin \left(A+B\right)}{\cos \left(A+B\right)} \\ \Rightarrow \tan \left(A+B\right)=\frac{5}{2} \end{gathered}$$

Since, $\sin \left(A-B\right)=\frac{1}{\sqrt{10}}$ and $\cos \left(A-B\right)=\frac{3}{\sqrt{10}}$.

So,

$$\begin{gathered} \tan \left(A-B\right)=\frac{\sin \left(A-B\right)}{\cos \left(A-B\right)} \\ \Rightarrow \tan \left(A-B\right)=\frac{1}{3} \end{gathered}$$

Step 3: Find the required value.

$$\begin{gathered} \tan \left(2A\right)=\tan \left(\left(A+B\right)+\left(A-B\right)\right) \\ \Rightarrow =\frac{\left({\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.)}+{\displaystyle (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}{1-\left({\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}\left[\because \tan \left(A+B\right)=\frac{\tan \left(A\right)+\tan \left(B\right)}{1-\tan \left(A\right)\tan \left(B\right)}\right] \\ \Rightarrow =17 \end{gathered}$$

Hence, the required value is $17$.