Question

If \[0 < \alpha , \beta < 4 \pi ,cos \left( { \alpha + \beta } \right) = 54,sin \left( { \alpha - \beta } \right) = 135, \] then \[tan2 \alpha \] =
A. \[ \dfrac{{33}}{{56}} \]
B. \[ \dfrac{{56}}{{33}} \]
C. $ \dfrac{{16}}{{33}} $
D.None

Answer 1

Hint: To answer the value of \[tan2 \alpha \] we need to find the value of $ \sin 2 \alpha $ and $ \cos 2 \alpha $ such that we can find the value of \[tan2 \alpha \] . To find the value of $ \sin 2 \alpha $ and $ \cos 2 \alpha $ we need to find the value of $ \sin ( \alpha + \beta ) $ and $ \cos ( \alpha - \beta ) $ . Once we find the value of these values use angle sum formula and find the required answer.

Complete step-by-step answer:
Given
 \[cos \left( { \alpha + \beta } \right) = 54,{ \text{ }}sin \left( { \alpha - \beta } \right) = 135 \]
Now aplly the formula that the sum of squares of sin and cos is always equal to 1 for the same value of angle.
So we get,
 \[\Rightarrow sin \left( { \alpha + \beta } \right) = \sqrt {1 - co{s^2} \left( { \alpha + \beta } \right)} = \sqrt {1 - {{ \left( { \dfrac{4}{5}} \right)}^2}} = \dfrac{3}{5} \]
Similarly,
 \[\Rightarrow cos \left( { \alpha - \beta } \right) = \sqrt {1 - si{n^2} \left( { \alpha - \beta } \right)} = \sqrt {1 - {{ \left( { \dfrac{{5}}{{13}}} \right)}^2}} = \dfrac{{12}}{{13}} \]
Now applying the formula of $ \sin 2 \alpha $
We get,
 \[\Rightarrow sin \left( {2 \alpha } \right) = sin \left( { \alpha + \beta + \alpha - \beta } \right) = sin \left( { \alpha + \beta } \right)cos \left( { \alpha - \beta } \right) + sin \left( { \alpha - \beta } \right)cos \left( { \alpha + \beta } \right) \]
On putting the given value we get,
 \[ = \dfrac{3}{5} \times \dfrac{{12}}{{13}} + \dfrac{5}{{13}} \times \dfrac{4}{5} = \dfrac{{56}}{{65}} \]
Similarly the value of $ \cos 2 \alpha $
We get,
 \[\Rightarrow cos \left( {2 \alpha } \right) = cos \left( { \alpha + \beta + \alpha - \beta } \right) = cos \left( { \alpha + \beta } \right)cos \left( { \alpha - \beta } \right) - sin \left( { \alpha - \beta } \right)sin \left( { \alpha - \beta } \right) \]
On putting the given values we get,
 \[ = \dfrac{{4 \times 12}}{{5 \times 13}} - \dfrac{{5 \times 13}}{{13 \times 5}} = \dfrac{{33}}{{65}} \]
Now applying the formula \[tan2 \alpha \] in terms of sin and cos we get
 \[tan2 \alpha = \dfrac{{sin2 \alpha }}{{cos2 \alpha }} \]
On putting the above values of $ \sin 2 \alpha $ and $ \cos 2 \alpha $
We get,
 \[ = \dfrac{{56}}{{65}} \times \dfrac{{65}}{{33}} = \dfrac{{56}}{{33}} \]
Hence the value of \[tan2 \alpha \] is \[ \dfrac{{56}}{{33}} \]
So, the correct answer is “Option B”.

Note: In this question students should know the adjustment of angle like \[cos \left( {2 \alpha } \right) = cos \left( { \alpha + \beta + \alpha - \beta } \right) \] otherwise this problem can not be solved easily. Might have to face difficulty as there seems to be no other easy way to solve this.