Question

verify the following identity.
cot
2
−
cot x=csc x
which of the following four statements establishes the identity?
a.
cot
2
−
cot x=
1−cos x
sin x
−
1
sin x
=csc x
b.
cot
2
−
cot x=
1−cos x
sin x
−
cos x
sin x
=csc x
c.
cot
2
−
cot x=
1+cos x
sin x
−
1
sin x
=csc x
d.
cot
2
−
cot x=
1+cos x
sin x
−
cos x
sin x
=csc x

Explanation:

Step1: Use trigonometric identities

We know that \(\cot\frac{x}{2}=\frac{\sin x}{1 - \cos x}\) (half - angle formula: \(\cot\frac{\alpha}{2}=\frac{\sin\alpha}{1-\cos\alpha}\)) and \(\csc x=\frac{1}{\sin x}\).
Starting with the left - hand side of the identity \(\cot\frac{x}{2}-\cot x\).
First, \(\cot\frac{x}{2}=\frac{\sin x}{1 - \cos x}\) and \(\cot x=\frac{\cos x}{\sin x}\).

Step2: Find a common denominator

$$
LATEXBLOCK0
$$

Since \(\sin^{2}x+\cos^{2}x = 1\), we have:

$$
LATEXBLOCK1
$$

Answer:

A. \(\cot\frac{x}{2}-\cot x=\frac{1-\cos x}{\sin x}-\frac{\cos x}{\sin x}=\frac{1-\cos x-\cos x}{\sin x}=\frac{1 - 2\cos x}{\sin x}\) (Incorrect).
B. \(\cot\frac{x}{2}-\cot x=\frac{1-\cos x}{\sin x}-\frac{\cos x}{\sin x}=\frac{1-\cos x-\cos x}{\sin x}\) (Incorrect).
C. \(\cot\frac{x}{2}-\cot x=\frac{1+\cos x}{\sin x}-\frac{\cos x}{\sin x}=\frac{1+\cos x-\cos x}{\sin x}=\frac{1}{\sin x}=\csc x\) (Correct).
D. \(\cot\frac{x}{2}-\cot x=\frac{1+\cos x}{\sin x}-\frac{\cos x}{\sin x}=\frac{1+\cos x-\cos x}{\sin x}\) (Incorrect).

So the answer is C.