Hello! Verify the identity. Please show your work! Use trigonometric identities to verify each expression is equal.


Yes this is from edge.

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.

pls help me..... the question is in the attachment.... thank you


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