Answer:
See Below.
Step-by-step explanation:
We want to verify the identity:
[tex]\displaystyle \csc^2 x -2\csc x \cot x +\cot ^2 x = \tan^2\left(\frac{x}{2}\right)[/tex]
Note that the left-hand side is a perfect square trinomial pattern. Namely:
[tex]a^2-2ab+b^2=(a-b)^2[/tex]
If we let a = csc(x) and b = cot(x), we can factor it as such:
[tex]\displaystyle (\csc x - \cot x)^2 = \tan^2\left(\frac{x}{2}\right)[/tex]
Let csc(x) = 1 / sin(x) and cot(x) = cos(x) / sin(x):
[tex]\displaystyle \left(\frac{1}{\sin x}-\frac{\cos x }{\sin x}\right)^2=\tan^2\left(\frac{x}{2}\right)[/tex]
Combine fractions:
[tex]\displaystyle \left(\frac{1-\cos x}{\sin x}\right)^2=\tan^2\left(\frac{x}{2}\right)[/tex]
Square (but do not simplify yet):
[tex]\displaystyle \frac{(1-\cos x)^2}{\sin ^2x}=\tan^2\left(\frac{x}{2}\right)[/tex]
Now, we can make a substitution. Let u = x / 2. So, x = 2u. Substitute:
[tex]\displaystyle \frac{(1-\cos 2u)^2}{\sin ^22u}=\tan^2u[/tex]
Recall that cos(2u) = 1 - sin²(u). Hence:
[tex]\displaystyle \frac{(1-(1-2\sin^2u))^2}{\sin ^2 2u}=\tan^2u[/tex]
Simplify:
[tex]\displaystyle \frac{4\sin^4 u}{\sin ^2 2u}=\tan^2 u[/tex]
Recall that sin(2u) = 2sin(u)cos(u). Hence:
[tex]\displaystyle \frac{4\sin^4 u}{(2\sin u\cos u)^2}=\tan^2 u[/tex]
Square:
[tex]\displaystyle \frac{4\sin^4 u}{4\sin^2 u\cos ^2u}=\tan^2 u[/tex]
Cancel:
[tex]\displaystyle \frac{\sin ^2 u}{\cos ^2 u}=\tan ^2 u[/tex]
Since sin(u) / cos(u) = tan(u):
[tex]\displaystyle \left(\frac{\sin u}{\cos u}\right)^2=\tan^2u=\tan^2u[/tex]
We can substitute u back for x / 2:
[tex]\displaystyle \tan^2\left(\frac{x}{2}\right)= \tan^2\left(\frac{x}{2}\right)[/tex]
Hence proven.
