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Question: If \[0 < \alpha , \beta < 4 \pi ,cos \left( { \alpha + \beta } \right) = 54,sin \left( { \alpha - \b...

If 0<α,β<4π,cos(α+β)=54,sin(α−β)=135,0 < \alpha , \beta < 4 \pi ,cos \left( { \alpha + \beta } \right) = 54,sin \left( { \alpha - \beta } \right) = 135, then tan2αtan2 \alpha =
A. 3356\dfrac{{33}}{{56}}
B. 5633\dfrac{{56}}{{33}}
C. 1633\dfrac{{16}}{{33}}
D.None

Explanation

Solution

Hint : To answer the value of tan2αtan2 \alpha we need to find the value of sin⁡2α\sin 2 \alpha and cos⁡2α\cos 2 \alpha such that we can find the value of tan2αtan2 \alpha . To find the value of sin⁡2α\sin 2 \alpha and cos⁡2α\cos 2 \alpha we need to find the value of sin⁡(α+β)\sin ( \alpha + \beta ) and cos⁡(α−β)\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(α+β)=54, sin(α−β)=135cos \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,
⇒sin(α+β)=1−cos2(α+β)=1−(45)2=35\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,
⇒cos(α−β)=1−sin2(α−β)=1−(513)2=1213\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⁡2α\sin 2 \alpha
We get,
⇒sin(2α)=sin(α+β+α−β)=sin(α+β)cos(α−β)+sin(α−β)cos(α+β)\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,
=35×1213+513×45=5665= \dfrac{3}{5} \times \dfrac{{12}}{{13}} + \dfrac{5}{{13}} \times \dfrac{4}{5} = \dfrac{{56}}{{65}}
Similarly the value of cos⁡2α\cos 2 \alpha
We get,
⇒cos(2α)=cos(α+β+α−β)=cos(α+β)cos(α−β)−sin(α−β)sin(α−β)\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,
=4×125×13−5×1313×5=3365= \dfrac{{4 \times 12}}{{5 \times 13}} - \dfrac{{5 \times 13}}{{13 \times 5}} = \dfrac{{33}}{{65}}
Now applying the formula tan2αtan2 \alpha in terms of sin and cos we get
tan2α=sin2αcos2αtan2 \alpha = \dfrac{{sin2 \alpha }}{{cos2 \alpha }}
On putting the above values of sin⁡2α\sin 2 \alpha and cos⁡2α\cos 2 \alpha
We get,
=5665×6533=5633= \dfrac{{56}}{{65}} \times \dfrac{{65}}{{33}} = \dfrac{{56}}{{33}}
Hence the value of tan2αtan2 \alpha is 5633\dfrac{{56}}{{33}}
So, the correct answer is “Option B”.

Note : In this question students should know the adjustment of angle like cos(2α)=cos(α+β+α−β)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.