Answer:
See prove of the identity below
Step-by-step explanation:
Recall that the following difference can be factored out as:
[tex](1-tan^4(A)) = (1+tan^2(A)) *(1-tan^2(A))[/tex]
Now we look at the right side of the equal sign, and recall the identity that relates the sine of a double angle 2A with an expression in terms of the tan(A):
[tex]sin(2A) = \frac{2\,tan(A)}{1+tan^2(A)}[/tex]
and similarly, the tangent of the double angle "2A" can be written as:
[tex]tan (2A) = \frac{2*tan(A)}{1-tan^2(A)}[/tex]
We can then combine the two expressions on the right using the common denominator: [tex](1-tan^4(A)) = (1+tan^2(A)) *(1-tan^2(A))[/tex]:
[tex]tan(2A) +sin(2A)=\frac{2*tan(A)*(1+tan^2(A))}{1-tan^4(A)} +\frac{2*tan(A)*(1-tan^2(A))}{1-tan^4(A)} \\=\frac{2*tan(A)+2*tan(A)}{1-tan^4(A)} =\frac{4*tan(A)}{1-tan^4(A)}[/tex]
And we have proved the identity since the right hand side becomes exactly equal to the left hand side of the equal sign.
