Question

If we have trigonometric ratios $\cos \left( \alpha +\beta \right)=\dfrac{4}{5},\sin \left( \alpha -\beta \right)=\dfrac{5}{13}$ and $\alpha ,\beta$ lie between 0 and $\dfrac{\pi }{4}$, find the value of $\tan 2\alpha$.

Answer 1

Given that $\alpha ,\beta$ lie between 0 and $\dfrac{\pi }{4}$, then $\left( \alpha +\beta \right)$ lie between 0 and $\dfrac{\pi }{2}$. As $\sin \left( \alpha -\beta \right)=\dfrac{5}{13}$ and $\alpha ,\beta$ lie between 0 and $\dfrac{\pi }{4}$, $\alpha \gg \beta$ and $\left( \alpha -\beta \right)$ lie between 0 and $\dfrac{\pi }{4}$.
Using the relation ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we can find the values of $\cos \left( \alpha -\beta \right),\sin \left( \alpha +\beta \right)$. The required term is $\tan 2\alpha$. By writing $2\alpha$as$\left( \alpha +\beta \right)+\left( \alpha -\beta \right)$, we get $\tan 2\alpha$ as $\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)$.
We know the formula of $\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\times \tan B}$. Using this formula we can get the answer.

Complete step-by-step solution:
In the question, it is given that $\cos \left( \alpha +\beta \right)=\dfrac{4}{5},\sin \left( \alpha -\beta \right)=\dfrac{5}{13}$.
We know the relation that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Using this equation to get the values of $\cos \left( \alpha -\beta \right),\sin \left( \alpha +\beta \right)$.

& {{\sin }^{2}}\left( \alpha +\beta \right)+{{\cos }^{2}}\left( \alpha +\beta \right)=1 \\\ & {{\sin }^{2}}\left( \alpha +\beta \right)=1-{{\cos }^{2}}\left( \alpha +\beta \right) \\\ & \sin \left( \alpha +\beta \right)=\sqrt{1-{{\cos }^{2}}\left( \alpha +\beta \right)} \\\ \end{aligned}$$ Substituting the value of $\cos \left( \alpha +\beta \right)=\dfrac{4}{5}$ in the above equation we get, $$\begin{aligned} & \sin \left( \alpha +\beta \right)=\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}} \\\ & \sin \left( \alpha +\beta \right)=\sqrt{1-\dfrac{16}{25}} \\\ & \sin \left( \alpha +\beta \right)=\sqrt{\dfrac{25-16}{25}} \\\ \end{aligned}$$ $$\sin \left( \alpha +\beta \right)=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}$$ We know that $$\tan \left( \alpha +\beta \right)=\dfrac{\sin \left( \alpha +\beta \right)}{\cos \left( \alpha +\beta \right)}$$ Substituting the values of $$\sin \left( \alpha +\beta \right)$$ and $$\cos \left( \alpha +\beta \right)$$ in the above equation, we get $$\tan \left( \alpha +\beta \right)=\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}=\dfrac{3}{4}\to \left( 1 \right)$$ Similarly, for $\left( \alpha -\beta \right)$, we get $$\begin{aligned} & {{\sin }^{2}}\left( \alpha -\beta \right)+{{\cos }^{2}}\left( \alpha -\beta \right)=1 \\\ & {{\cos }^{2}}\left( \alpha -\beta \right)=1-{{\sin }^{2}}\left( \alpha -\beta \right) \\\ & {{\cos }^{2}}\left( \alpha -\beta \right)=\sqrt{1-{{\sin }^{2}}\left( \alpha -\beta \right)} \\\ \end{aligned}$$ Substituting the value of $\sin \left( \alpha -\beta \right)=\dfrac{5}{13}$ in the above equation, we get $$\begin{aligned} & \cos \left( \alpha -\beta \right)=\sqrt{1-{{\left( \dfrac{5}{13} \right)}^{2}}} \\\ & \cos \left( \alpha -\beta \right)=\sqrt{1-\dfrac{25}{169}} \\\ & \cos \left( \alpha -\beta \right)=\sqrt{\dfrac{169-25}{169}} \\\ \end{aligned}$$ $$\cos \left( \alpha -\beta \right)=\sqrt{\dfrac{144}{169}}=\dfrac{12}{13}$$ We know that $$\tan \left( \alpha -\beta \right)=\dfrac{\sin \left( \alpha -\beta \right)}{\cos \left( \alpha -\beta \right)}$$ Substituting the values of $$\sin \left( \alpha -\beta \right)$$ and $$\cos \left( \alpha -\beta \right)$$ in the above equation, we get $$\tan \left( \alpha -\beta \right)=\dfrac{\dfrac{5}{13}}{\dfrac{12}{13}}=\dfrac{5}{12}\to \left( 2 \right)$$ The required term is $\tan 2\alpha$. To get the required term, we rewrite the expression as $\tan 2\alpha =\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)\to \left( 3 \right)$ We know the formula $\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\times \tan B}\to \left( 4 \right)$, using this formula in equation-3, we get $\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)=\dfrac{\tan \left( \alpha +\beta \right)+\tan \left( \alpha -\beta \right)}{1-\tan \left( \alpha +\beta \right)\tan \left( \alpha -\beta \right)}$. Using the values of $\tan \left( \alpha +\beta \right)\text{ and }\tan \left( \alpha -\beta \right)$from the equations - 1 and 2, we get, $\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)=\dfrac{\dfrac{3}{4}+\dfrac{5}{12}}{1-\dfrac{3}{4}\times \dfrac{5}{12}}$ By simplifying, we get $\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)=\dfrac{\dfrac{3\times 3+5}{12}}{\dfrac{4\times 12-15}{4\times 12}}=\dfrac{\dfrac{14}{12}}{\dfrac{48-15}{48}}=\dfrac{14}{12}\times \dfrac{48}{33}$ $\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)=\dfrac{56}{33}$ $\therefore \tan 2\alpha =\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)=\dfrac{56}{33}$. **Note:** An alternate way to do it is to convert both the given data into single trigonometric function and applying the inverse trigonometric concepts in it. That is $\begin{aligned} & \cos \left( \alpha +\beta \right)=\dfrac{4}{5}\Rightarrow \sin \left( \alpha +\beta \right)=\sqrt{1-{{\cos }^{2}}\left( \alpha +\beta \right)} \\\ & \Rightarrow \sin \left( \alpha +\beta \right)=\sqrt{1-\dfrac{16}{25}}=\dfrac{3}{5} \\\ & \sin \left( \alpha -\beta \right)=\dfrac{5}{13} \\\ \end{aligned}$ We can write that $\begin{aligned} & \alpha +\beta ={{\sin }^{-1}}\dfrac{3}{5} \\\ & \alpha -\beta ={{\sin }^{-1}}\dfrac{5}{13} \\\ \end{aligned}$ We know that ${{\sin }^{-1}}\dfrac{a}{b}={{\tan }^{-1}}\dfrac{a}{\sqrt{{{b}^{2}}-{{a}^{2}}}}$ From this, we can write $\begin{aligned} & \alpha +\beta ={{\sin }^{-1}}\dfrac{3}{5}={{\tan }^{-1}}\dfrac{3}{\sqrt{{{5}^{2}}-{{3}^{2}}}}={{\tan }^{-1}}\dfrac{3}{4} \\\ & \alpha -\beta ={{\sin }^{-1}}\dfrac{5}{13}={{\tan }^{-1}}\dfrac{5}{\sqrt{{{13}^{2}}-{{5}^{2}}}}={{\tan }^{-1}}\dfrac{5}{12} \\\ \end{aligned}$ Adding the two equations, we get $2\alpha ={{\tan }^{-1}}\dfrac{3}{4}+{{\tan }^{-1}}\dfrac{5}{12}$ Applying Tan on both sides and using equation - 4, we get $\tan 2\alpha =\tan \left( {{\tan }^{-1}}\dfrac{3}{4}+{{\tan }^{-1}}\dfrac{5}{12} \right)=\dfrac{\tan \left( {{\tan }^{-1}}\dfrac{3}{4} \right)+\tan \left( {{\tan }^{-1}}\dfrac{5}{12} \right)}{1-\tan \left( {{\tan }^{-1}}\dfrac{3}{4} \right)\times \tan \left( {{\tan }^{-1}}\dfrac{5}{12} \right)}=\dfrac{\dfrac{3}{4}+\dfrac{5}{12}}{1-\dfrac{3}{4}\times \dfrac{5}{12}}$ $\tan 2\alpha =\dfrac{56}{33}$ same answer as the one we got with the other process.