5.2 Prove the following identity:

[tex]\[
\frac{1}{\sin \left(180^{\circ} + x\right) \cdot \sin \left(360^{\circ} - x\right)} - \cot^2 \left(180^{\circ} - x\right) = 1
\][/tex]



Answer :

To prove the given trigonometric identity:
[tex]\[ \frac{1}{\sin \left(180^{\circ}+x\right) \cdot \sin \left(360^{\circ}-x\right)} - \cot^2\left(180^{\circ}-x\right) = 1 \][/tex]

Let's start using some fundamental trigonometric identities and simplify each term step-by-step.

### Step 1: Simplify [tex]\(\sin(180^\circ + x)\)[/tex]

We know the sine function has the identity:
[tex]\[ \sin(180^\circ + x) = -\sin(x) \][/tex]

### Step 2: Simplify [tex]\(\sin(360^\circ - x)\)[/tex]

Similarly, for the sine function:
[tex]\[ \sin(360^\circ - x) = \sin(-x) = -\sin(x) \][/tex]

### Step 3: Simplify [tex]\(\cot(180^\circ - x)\)[/tex]

Using the cotangent identity:
[tex]\[ \cot(180^\circ - x) = -\cot(x) \][/tex]

### Step 4: Substitute the simplified terms back into the equation

Substitute [tex]\(\sin(180^\circ + x), \sin(360^\circ - x),\)[/tex] and [tex]\(\cot(180^\circ - x)\)[/tex] into the original identity:

[tex]\[ \frac{1}{\left(-\sin(x)\right) \cdot \left(-\sin(x)\right)} - \left(-\cot(x)\right)^2 \][/tex]

Since [tex]\((-1) \cdot (-1) = 1\)[/tex], we have:

[tex]\[ \frac{1}{\sin(x) \cdot \sin(x)} - (\cot(x))^2 \][/tex]

### Step 5: Simplify the left-hand side

Rewrite the left-hand side:
[tex]\[ \frac{1}{\sin^2(x)} - \cot^2(x) \][/tex]

Recall that:
[tex]\[ \cot(x) = \frac{\cos(x)}{\sin(x)} \implies \cot^2(x) = \frac{\cos^2(x)}{\sin^2(x)} \][/tex]

Therefore, we can express the left-hand side as:
[tex]\[ \frac{1}{\sin^2(x)} - \frac{\cos^2(x)}{\sin^2(x)} \][/tex]

### Step 6: Combine the fractions

Combine the fractions under a common denominator:
[tex]\[ \frac{1 - \cos^2(x)}{\sin^2(x)} \][/tex]

### Step 7: Utilize a Pythagorean identity

Recall the Pythagorean identity:
[tex]\[ \sin^2(x) + \cos^2(x) = 1 \implies 1 - \cos^2(x) = \sin^2(x) \][/tex]

Substitute [tex]\(1 - \cos^2(x)\)[/tex] with [tex]\(\sin^2(x)\)[/tex]:
[tex]\[ \frac{\sin^2(x)}{\sin^2(x)} = 1 \][/tex]

### Step 8: Conclusion

Thus, we have shown that:
[tex]\[ \frac{1}{\sin\left(180^\circ + x\right) \cdot \sin\left(360^\circ - x\right)} - \cot^2\left(180^\circ - x\right) = 1 \][/tex]

Hence, the given identity is proven to be true.