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 $\left( {\dfrac{\pi }{4}} \right)$, then find the value of $\tan \left( {2\alpha } \right)$.
(A) $\dfrac{{16}}{{63}}$
(B) $\dfrac{{56}}{{33}}$
(C) $\dfrac{{28}}{{33}}$
(D) None of these

Answer 1

Hint: The given question requires us to find the value of $\tan \left( {2\alpha } \right)$, when we are given the values of $\cos \left( {\alpha + \beta } \right)$ and \[\sin \left( {\alpha - \beta } \right)\]. We are also given that the angles $\alpha $ and $\beta $ lie in the range $0$ and $\left( {\dfrac{\pi }{4}} \right)$. So, this means that the angle $\left( {2\alpha } \right)$ lies between $0$ and $\left( {\dfrac{\pi }{2}} \right)$. So, the tangent of the angle must be positive as it lies in the first quadrant. We will use the compound formula of tangent to find the value of the trigonometric function for the multiple angle.

Complete answer:
So, we are given the value of trigonometric ratio $\cos \left( {\alpha + \beta } \right) = \dfrac{4}{5}$.
So, \[\cos \left( {\alpha + \beta } \right) = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}} = \dfrac{4}{5}\].
So, we know the ratio of Base and Hypotenuse. Let ${\text{Base = 4x}}$ and ${\text{Hypotenuse = 5x}}$.
Now, using Pythagoras theorem, we have,
${\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Altitude} \right)^2}$
$ \Rightarrow {\left( {5x} \right)^2} = {\left( {4x} \right)^2} + {\left( {Altitude} \right)^2}$
$ \Rightarrow {\left( {Altitude} \right)^2} = 25{x^2} - 16{x^2}$
$ \Rightarrow {\left( {Altitude} \right)^2} = 9{x^2}$
$ \Rightarrow Altitude = 3x$
Therefore calculating the tangent of the angle \[\left( {\alpha + \beta } \right)\] as:
\[\tan \left( {\alpha + \beta } \right) = \dfrac{{{\text{Altitude}}}}{{{\text{Base}}}} = \dfrac{{3x}}{{4x}} = \dfrac{3}{4}\]
So, we have, $\sin \left( {\alpha - \beta } \right) = \dfrac{5}{{13}}$
So, \[\sin \left( {\alpha - \beta } \right) = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}} = \dfrac{5}{{13}}\].
So, we know the ratio of Base and Hypotenuse. Let ${\text{Perpendicular = 5x}}$ and ${\text{Hypotenuse = 13x}}$.
Now, using Pythagoras theorem, we have,
${\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Altitude} \right)^2}$
$ \Rightarrow {\left( {13x} \right)^2} = {\left( {Base} \right)^2} + {\left( {5x} \right)^2}$
$ \Rightarrow {\left( {Base} \right)^2} = 169{x^2} - 25{x^2}$
$ \Rightarrow {\left( {Base} \right)^2} = 144{x^2}$
$ \Rightarrow Base = 12x$
Therefore calculating the tangent of the angle \[\left( {\alpha - \beta } \right)\] as:
\[\tan \left( {\alpha - \beta } \right) = \dfrac{{{\text{Altitude}}}}{{{\text{Base}}}} = \dfrac{{5x}}{{12x}} = \dfrac{5}{{12}}\]
Now, we have to find the value of $\tan \left( {2\alpha } \right)$.
So, we know the compound angle formula of tangent as $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$.
So, we get, $\tan \left( {2\alpha } \right) = \tan \left[ {\left( {\alpha + \beta } \right) + \left( {\alpha - \beta } \right)} \right]$
$ \Rightarrow \tan \left( {2\alpha } \right) = \dfrac{{\tan \left( {\alpha + \beta } \right) + \tan \left( {\alpha - \beta } \right)}}{{1 - \tan \left( {\alpha + \beta } \right)\tan \left( {\alpha - \beta } \right)}}$
Substituting the values of \[\tan \left( {\alpha - \beta } \right)\] and \[\tan \left( {\alpha + \beta } \right)\], we get,
$ \Rightarrow \tan \left( {2\alpha } \right) = \dfrac{{\dfrac{3}{4} + \dfrac{5}{{12}}}}{{1 - \dfrac{3}{4} \times \dfrac{5}{{12}}}}$
Taking the LCM of the denominators, we get,
$ \Rightarrow \tan \left( {2\alpha } \right) = \dfrac{{\dfrac{9}{{12}} + \dfrac{5}{{12}}}}{{1 - \dfrac{{15}}{{48}}}}$
$ \Rightarrow \tan \left( {2\alpha } \right) = \dfrac{{\dfrac{{9 + 5}}{{12}}}}{{\dfrac{{48 - 15}}{{48}}}}$
Simplifying the expression, we get,
$ \Rightarrow \tan \left( {2\alpha } \right) = \dfrac{{\left( {\dfrac{{14}}{{12}}} \right)}}{{\left( {\dfrac{{33}}{{48}}} \right)}}$
$ \Rightarrow \tan \left( {2\alpha } \right) = \left( {\dfrac{{14 \times 48}}{{12 \times 33}}} \right)$
Computing the product of numbers,
$ \Rightarrow \tan \left( {2\alpha } \right) = \left( {\dfrac{{14 \times 4}}{{33}}} \right)$
$ \Rightarrow \tan \left( {2\alpha } \right) = \left( {\dfrac{{56}}{{33}}} \right)$
So, we get the value of $\tan \left( {2\alpha } \right)$ as $\left( {\dfrac{{56}}{{33}}} \right)$.

Note:
We must remember the formulae for finding the values of trigonometric functions for compound angles such as $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$ and $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$. We should know the method of finding values of trigonometric functions when we are given the value of any one of them. One must take care of calculations so as to be sure of the final answer.