Question

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

A. \[\dfrac{{16}}{{63}}\]
B. \[\dfrac{{56}}{{33}}\]
C. \[\dfrac{{28}}{{33}}\]
D. None of these

Answer 1

Hint: In this question, for determining the value of \[\tan 2\alpha \], we need to express \[\tan 2\alpha \] in terms of \[\tan \left( {\alpha + \beta + \alpha - \beta } \right)\]. For this, we have to find the values of \[\cos \left( {\alpha - \beta } \right)\]and \[\sin \left( {\alpha - \beta } \right)\] from the given data.


Complete step by step answer:
We know that \[\cos \left( {\alpha + \beta } \right) = \dfrac{4}{5}\].
But \[\sin x = \sqrt {1 - {{\cos }^2}x} \].
Thus, we get
\[\sin \left( {\alpha + \beta } \right) = \sqrt {1 - {{\cos }^2}\left( {\alpha + \beta } \right)} \]
\[\sin \left( {\alpha + \beta } \right) = \sqrt {1 - {{\left( {\dfrac{4}{5}} \right)}^2}} \]
\[\sin \left( {\alpha + \beta } \right) = \dfrac{3}{5}\]

 Also, we know that \[\sin \left( {\alpha - \beta } \right) = \dfrac{5}{{13}}\]
But \[\cos x = \sqrt {1 - {{\sin }^2}x} \]
So, we get
\[\cos \left( {\alpha - \beta } \right) = \sqrt {1 - {{\sin }^2}\left( {\alpha - \beta } \right)} \]
\[\cos \left( {\alpha - \beta } \right) = \sqrt {1 - {{\left( {5/13} \right)}^2}} \]
\[\cos \left( {\alpha - \beta } \right) = \dfrac{{12}}{{13}}\]
 Therefore, we can say that \[\tan 2\alpha = \tan \left( {\alpha + \beta + \alpha - \beta } \right)\]
We know that \[\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a \times \tan b}}\]
So, \[\tan \left( {2\alpha } \right) = \dfrac{{\tan \left( {\alpha + \beta } \right) + \tan \left( {\alpha - \beta } \right)}}{{1 - \tan \left( {\alpha + \beta } \right) \times \tan \left( {\alpha - \beta } \right)}}\]
Now, let us express in terms of sin and cos
\[\tan \left( {2\alpha } \right) = \dfrac{{\dfrac{{\sin \left( {\alpha + \beta } \right)}}{{\cos \left( {\alpha + \beta } \right)}} + \dfrac{{\sin \left( {\alpha - \beta } \right)}}{{\cos \left( {\alpha - \beta } \right)}}}}{{1 - \dfrac{{\sin \left( {\alpha + \beta } \right)}}{{\cos \left( {\alpha + \beta } \right)}} \times \dfrac{{\sin \left( {\alpha - \beta } \right)}}{{\cos \left( {\alpha - \beta } \right)}}}}\]
Thus, we can have
\[\tan \left( {2\alpha } \right) = \dfrac{{\dfrac{{3/5}}{{4/5}} + \dfrac{{5/13}}{{12/13}}}}{{1 - \dfrac{{3/5}}{{4/5}} \times \dfrac{{5/13}}{{12/13}}}}\]
By simplifying, we get
\[\tan \left( {2\alpha } \right) = \dfrac{{56}}{{33}}\]
Hence, the value of \[\tan \left( {2\alpha } \right)\] is \[\dfrac{{56}}{{33}}\].


Therefore, the option (B) is correct.

Note: Many students make mistakes in solving the calculation part and applying trigonometric identities. This is the only way, through which we can solve the example in the simplest way. Converting all the trigonometric ratios into sine and cosine is necessary for solving trigonometric problems as this makes them simpler.