Answer:
Step-by-step explanation:
These are really difficult, but one thing I noticed pretty much across the board is that we state our trig ratios in terms of sin and cos using the Pythagorean identity:
[tex]sin^2x+cos^2x=1[/tex], even this one where tangents are involved. Let's begin:
If tan⁻1(x) = u (this is just a substitution to make things easier down the road), then tan(u) = x
If tan⁻1(y) = v, then tan(v) = y.
If that be the case, what we have as a rewrite from our original problem is
sin(u + v) which, according to the identity is:
[tex]sin(u-v)=sin(u)cos(v)-sin(v)cos(u)[/tex] which also tells us that we need those tangents in terms of sines and cosines, as stated earlier. We will take the Pythagorean identity and solve it for both sinx and cosx. Sinx first:
If [tex]cos^2x+sin^2x=1[/tex] then solving for sin:
[tex]\frac{cos^2x}{sin^2x}+\frac{sin^2x}{sin^2x}=\frac{1}{sin^2x}[/tex] (as you can see, I just divided everything by sin-squared):
[tex]cot^2x+1=\frac{1}{sin^2x}[/tex] and
[tex]\frac{1}{tan^2x}+1=\frac{1}{sin^2x}[/tex] and work on eliminating the denominators by multiplying through by sin-squared times tan-squared:
[tex]sin^2x*tan^2x(\frac{1}{tan^2x}+1=\frac{1}{sin^2x})[/tex] which simplifies to
[tex]sin^2x+sin^2x*tan^2x=tan^2x[/tex] and factor out on the left:
[tex]sin^2x(1+tan^2x)=tan^2x[/tex] and solve for sin-squared:
[tex]sin^2x=\frac{tan^2x}{1+tan^2x}[/tex] and take the square root of both sides:
[tex]sinx=\frac{tanx}{\sqrt{1+tan^2x} }[/tex] . Do the same to solve for cos(x) to get:
[tex]cosx=\frac{1}{\sqrt{1+tan^2x} }[/tex] and now we're ready to fill in our formula for sin(u + v):
[tex]\frac{tanu}{\sqrt{1+tan^2u} }*\frac{1}{\sqrt{1+tan^2v} }[/tex] - [tex]\frac{tanv}{\sqrt{1+tan^2} }*\frac{1}{\sqrt{1+tan^2u} }[/tex] and...
since tanu = x and tanv = y:
[tex]\frac{x}{\sqrt{1+x^2}}*\frac{1}{\sqrt{1+y^2}}-\frac{y}{\sqrt{1+y^2}}*\frac{1}{\sqrt{1+x^2} }[/tex] which, in the end, simplifies to
[tex]\frac{x-y}{\sqrt{1+x^2}*\sqrt{1+y^2} }[/tex] and you're done! Yikes to that!!!!
