Question 628525
{{{(2tan(alpha))/(1+tan^2(alpha))=sin(2*alpha)}}}
When you don't see any better solutions, it is a good idea to rewrite any tan's, cot's, sec's and csc's in terms of sin and/or cos. Since I do not yet see a faster solution, this is what I will do:
{{{(2(sin(alpha)/cos(alpha)))/(1+(sin(alpha)/cos(alpha))^2)=sin(2*alpha)}}}
TO get rid of the fractions within the fraction, we multiply the numerator and denominator of the "big" fraction by the least common denominator (LCD) of all the "little" fractions. THe LCD here is {{{(cos(alpha))^2}}}:
{{{((2(sin(alpha)/cos(alpha)))/(1+(sin(alpha)/cos(alpha))^2))((cos(alpha))^2/(cos(alpha))^2)=sin(2*alpha)}}}
In the numerator, one {{{cos(alpha) }}} cancels. In the denominator we use the Distributive property:
{{{(2sin(alpha)cos(alpha))/((cos(alpha))^2 + (sin(alpha))^2) = sin(2*alpha)}}}
We should now recognize both the numerator and denominator:
{{{sin(2*alpha)/1 = sin(2*alpha)}}}
which simplifies to
{{{sin(2*alpha) = sin(2*alpha)}}}
And we are finished!<br>
The second problem is impossible to solve without knowing what quadrants alpha and beta terminate in.