[tex]\tan(2\alpha)=\dfrac{2\tan \alpha }{1-\tan^2 \alpha}\\\\\\\dfrac{\sin(2\alpha)}{\cos(2\alpha)}=\dfrac{\dfrac{2\sin\alpha}{\cos\alpha}}{1-\dfrac{\sin ^2\alpha}{\cos^2\alpha}}\\\\\\\dfrac{\sin(2\alpha)}{\cos(2\alpha)}=\dfrac{\dfrac{2\sin\alpha}{\cos\alpha} }{\dfrac{\cos^2\alpha}{\cos^2\alpha}-\dfrac{\sin ^2\alpha}{\cos^2\alpha}}\\\\\\\dfrac{\sin(2\alpha)}{\cos(2\alpha)}=\dfrac{\dfrac{2\sin\alpha}{\cos\alpha} }{\dfrac{\cos^2\alpha-\sin^2 \alpha}{\cos^2\alpha}}[/tex]
[tex]\dfrac{\sin(2\alpha)}{\cos(2\alpha)}=\dfrac{2\sin\alpha}{\cos\alpha} \cdot\dfrac{\cos^2\alpha}{\cos^2\alpha-\sin^2 \alpha}\\\\\\\dfrac{\sin(2\alpha)}{\cos(2\alpha)}=\dfrac{2\sin\alpha\cos\alpha}{\cos^2\alpha-\sin^2 \alpha} \\\\\\\dfrac{\sin(2\alpha)}{\cos(2\alpha)}=\dfrac{\sin(2\alpha)}{\cos(2\alpha)}[/tex]
Answer:
See below.
Step-by-step explanation:
tan 2A = sin 2A / cos 2A
= 2 sinA cosA / (cos^2A - sin^2A)
Now divide top and bottom of the fraction by cos^2 A:
2 sinA cosA cos^2A sin^2 A
------------------- / ----------- - -------------
cos^2A cos^2 A cos^2 A
= 2 tan A / 1 - tan^2A).
