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
Verified581.1k+ views
Hint: 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 $\cos \left( \alpha -\beta \right),\sin \left( \alpha +\beta \right)$.
\[\begin{align}
& {{\sin }^{2}}\left( \alpha +\beta \right)+{{\cos }^{2}}\left( \alpha +\beta \right)=1 \\
& {{\sin }^{2}}\left( \alpha +\beta \right)=1-{{\cos }^{2}}\left( \alpha +\beta \right) \\
& \sin \left( \alpha +\beta \right)=\sqrt{1-{{\cos }^{2}}\left( \alpha +\beta \right)} \\
\end{align}\]
Substituting the value of $\cos \left( \alpha +\beta \right)=\dfrac{4}{5}$ in the above equation we get,
\[\begin{align}
& \sin \left( \alpha +\beta \right)=\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}} \\
& \sin \left( \alpha +\beta \right)=\sqrt{1-\dfrac{16}{25}} \\
& \sin \left( \alpha +\beta \right)=\sqrt{\dfrac{25-16}{25}} \\
\end{align}\]
\[\sin \left( \alpha +\beta \right)=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\]
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
\[\begin{align}
& {{\sin }^{2}}\left( \alpha -\beta \right)+{{\cos }^{2}}\left( \alpha -\beta \right)=1 \\
& {{\cos }^{2}}\left( \alpha -\beta \right)=1-{{\sin }^{2}}\left( \alpha -\beta \right) \\
& {{\cos }^{2}}\left( \alpha -\beta \right)=\sqrt{1-{{\sin }^{2}}\left( \alpha -\beta \right)} \\
\end{align}\]
Substituting the value of $\sin \left( \alpha -\beta \right)=\dfrac{5}{13}$ in the above equation, we get
\[\begin{align}
& \cos \left( \alpha -\beta \right)=\sqrt{1-{{\left( \dfrac{5}{13} \right)}^{2}}} \\
& \cos \left( \alpha -\beta \right)=\sqrt{1-\dfrac{25}{169}} \\
& \cos \left( \alpha -\beta \right)=\sqrt{\dfrac{169-25}{169}} \\
\end{align}\]
\[\cos \left( \alpha -\beta \right)=\sqrt{\dfrac{144}{169}}=\dfrac{12}{13}\]
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
$\begin{align}
& \cos \left( \alpha +\beta \right)=\dfrac{4}{5}\Rightarrow \sin \left( \alpha +\beta \right)=\sqrt{1-{{\cos }^{2}}\left( \alpha +\beta \right)} \\
& \Rightarrow \sin \left( \alpha +\beta \right)=\sqrt{1-\dfrac{16}{25}}=\dfrac{3}{5} \\
& \sin \left( \alpha -\beta \right)=\dfrac{5}{13} \\
\end{align}$
We can write that
$\begin{align}
& \alpha +\beta ={{\sin }^{-1}}\dfrac{3}{5} \\
& \alpha -\beta ={{\sin }^{-1}}\dfrac{5}{13} \\
\end{align}$
We know that ${{\sin }^{-1}}\dfrac{a}{b}={{\tan }^{-1}}\dfrac{a}{\sqrt{{{b}^{2}}-{{a}^{2}}}}$
From this, we can write
$\begin{align}
& \alpha +\beta ={{\sin }^{-1}}\dfrac{3}{5}={{\tan }^{-1}}\dfrac{3}{\sqrt{{{5}^{2}}-{{3}^{2}}}}={{\tan }^{-1}}\dfrac{3}{4} \\
& \alpha -\beta ={{\sin }^{-1}}\dfrac{5}{13}={{\tan }^{-1}}\dfrac{5}{\sqrt{{{13}^{2}}-{{5}^{2}}}}={{\tan }^{-1}}\dfrac{5}{12} \\
\end{align}$
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.
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 $\cos \left( \alpha -\beta \right),\sin \left( \alpha +\beta \right)$.
\[\begin{align}
& {{\sin }^{2}}\left( \alpha +\beta \right)+{{\cos }^{2}}\left( \alpha +\beta \right)=1 \\
& {{\sin }^{2}}\left( \alpha +\beta \right)=1-{{\cos }^{2}}\left( \alpha +\beta \right) \\
& \sin \left( \alpha +\beta \right)=\sqrt{1-{{\cos }^{2}}\left( \alpha +\beta \right)} \\
\end{align}\]
Substituting the value of $\cos \left( \alpha +\beta \right)=\dfrac{4}{5}$ in the above equation we get,
\[\begin{align}
& \sin \left( \alpha +\beta \right)=\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}} \\
& \sin \left( \alpha +\beta \right)=\sqrt{1-\dfrac{16}{25}} \\
& \sin \left( \alpha +\beta \right)=\sqrt{\dfrac{25-16}{25}} \\
\end{align}\]
\[\sin \left( \alpha +\beta \right)=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\]
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
\[\begin{align}
& {{\sin }^{2}}\left( \alpha -\beta \right)+{{\cos }^{2}}\left( \alpha -\beta \right)=1 \\
& {{\cos }^{2}}\left( \alpha -\beta \right)=1-{{\sin }^{2}}\left( \alpha -\beta \right) \\
& {{\cos }^{2}}\left( \alpha -\beta \right)=\sqrt{1-{{\sin }^{2}}\left( \alpha -\beta \right)} \\
\end{align}\]
Substituting the value of $\sin \left( \alpha -\beta \right)=\dfrac{5}{13}$ in the above equation, we get
\[\begin{align}
& \cos \left( \alpha -\beta \right)=\sqrt{1-{{\left( \dfrac{5}{13} \right)}^{2}}} \\
& \cos \left( \alpha -\beta \right)=\sqrt{1-\dfrac{25}{169}} \\
& \cos \left( \alpha -\beta \right)=\sqrt{\dfrac{169-25}{169}} \\
\end{align}\]
\[\cos \left( \alpha -\beta \right)=\sqrt{\dfrac{144}{169}}=\dfrac{12}{13}\]
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
$\begin{align}
& \cos \left( \alpha +\beta \right)=\dfrac{4}{5}\Rightarrow \sin \left( \alpha +\beta \right)=\sqrt{1-{{\cos }^{2}}\left( \alpha +\beta \right)} \\
& \Rightarrow \sin \left( \alpha +\beta \right)=\sqrt{1-\dfrac{16}{25}}=\dfrac{3}{5} \\
& \sin \left( \alpha -\beta \right)=\dfrac{5}{13} \\
\end{align}$
We can write that
$\begin{align}
& \alpha +\beta ={{\sin }^{-1}}\dfrac{3}{5} \\
& \alpha -\beta ={{\sin }^{-1}}\dfrac{5}{13} \\
\end{align}$
We know that ${{\sin }^{-1}}\dfrac{a}{b}={{\tan }^{-1}}\dfrac{a}{\sqrt{{{b}^{2}}-{{a}^{2}}}}$
From this, we can write
$\begin{align}
& \alpha +\beta ={{\sin }^{-1}}\dfrac{3}{5}={{\tan }^{-1}}\dfrac{3}{\sqrt{{{5}^{2}}-{{3}^{2}}}}={{\tan }^{-1}}\dfrac{3}{4} \\
& \alpha -\beta ={{\sin }^{-1}}\dfrac{5}{13}={{\tan }^{-1}}\dfrac{5}{\sqrt{{{13}^{2}}-{{5}^{2}}}}={{\tan }^{-1}}\dfrac{5}{12} \\
\end{align}$
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.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success
Master Class 11 English: Engaging Questions & Answers for Success
Master Class 11 Social Science: Engaging Questions & Answers for Success
Master Class 11 Biology: Engaging Questions & Answers for Success
Class 11 Question and Answer - Your Ultimate Solutions Guide
Master Class 11 Business Studies: Engaging Questions & Answers for Success
Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE
10 examples of friction in our daily life
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Difference Between Prokaryotic Cells and Eukaryotic Cells
What are Quantum numbers Explain the quantum number class 11 chemistry CBSE
1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE