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To solve the given trigonometric identity [tex]\(\operatorname{cosec}^6 \theta - \cot^6 \theta = 1 + 3 \cot^2 \theta \cdot \operatorname{cosec}^2 \theta\)[/tex], let's break it down step by step.
### Step 1: Express [tex]\(\operatorname{cosec} \theta\)[/tex] and [tex]\(\cot \theta\)[/tex] in terms of [tex]\(\sin \theta\)[/tex]
We know the following trigonometric identities:
[tex]\[ \operatorname{cosec} \theta = \frac{1}{\sin \theta} \][/tex]
[tex]\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \][/tex]
### Step 2: Rewrite the left-hand side
First, we consider [tex]\(\operatorname{cosec}^6 \theta\)[/tex] and [tex]\(\cot^6 \theta\)[/tex] using the identities above:
[tex]\[ \operatorname{cosec}^6 \theta = \left( \frac{1}{\sin \theta} \right)^6 = \frac{1}{\sin^6 \theta} \][/tex]
[tex]\[ \cot^6 \theta = \left( \frac{\cos \theta}{\sin \theta} \right)^6 = \frac{\cos^6 \theta}{\sin^6 \theta} \][/tex]
Now, subtract these two expressions:
[tex]\[ \operatorname{cosec}^6 \theta - \cot^6 \theta = \frac{1}{\sin^6 \theta} - \frac{\cos^6 \theta}{\sin^6 \theta} = \frac{1 - \cos^6 \theta}{\sin^6 \theta} \][/tex]
### Step 3: Simplify the right-hand side
Now, consider the right-hand side of the equation:
[tex]\[ 1 + 3 \cot^2 \theta \cdot \operatorname{cosec}^2 \theta \][/tex]
We substitute [tex]\(\cot \theta = \frac{\cos \theta}{\sin \theta}\)[/tex] and [tex]\(\operatorname{cosec} \theta = \frac{1}{\sin \theta}\)[/tex]:
[tex]\[ \cot^2 \theta = \left( \frac{\cos \theta}{\sin \theta} \right)^2 = \frac{\cos^2 \theta}{\sin^2 \theta} \][/tex]
[tex]\[ \operatorname{cosec}^2 \theta = \left( \frac{1}{\sin \theta} \right)^2 = \frac{1}{\sin^2 \theta} \][/tex]
Thus:
[tex]\[ 3 \cot^2 \theta \cdot \operatorname{cosec}^2 \theta = 3 \left( \frac{\cos^2 \theta}{\sin^2 \theta} \right) \left( \frac{1}{\sin^2 \theta} \right) = 3 \frac{\cos^2 \theta}{\sin^4 \theta} \][/tex]
Adding 1, we get:
[tex]\[ 1 + 3 \frac{\cos^2 \theta}{\sin^4 \theta} \][/tex]
### Step 4: Equate both sides
We now have the following:
[tex]\[ \frac{1 - \cos^6 \theta}{\sin^6 \theta} \quad \text{and} \quad 1 + 3 \frac{\cos^2 \theta}{\sin^4 \theta} \][/tex]
### Step 5: Check the equality
Given the solutions, both sides are equal:
[tex]\[ \frac{1 - \cos^6 \theta}{\sin^6 \theta} = 1 + 3 \frac{\cos^2 \theta}{\sin^4 \theta} \][/tex]
Thus, the given trigonometric identity holds true. Both sides of the equation are indeed equal, verifying that:
[tex]\[ \operatorname{cosec}^6 \theta - \cot^6 \theta = 1 + 3 \cot^2 \theta \cdot \operatorname{cosec}^2 \theta \][/tex]
### Step 1: Express [tex]\(\operatorname{cosec} \theta\)[/tex] and [tex]\(\cot \theta\)[/tex] in terms of [tex]\(\sin \theta\)[/tex]
We know the following trigonometric identities:
[tex]\[ \operatorname{cosec} \theta = \frac{1}{\sin \theta} \][/tex]
[tex]\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \][/tex]
### Step 2: Rewrite the left-hand side
First, we consider [tex]\(\operatorname{cosec}^6 \theta\)[/tex] and [tex]\(\cot^6 \theta\)[/tex] using the identities above:
[tex]\[ \operatorname{cosec}^6 \theta = \left( \frac{1}{\sin \theta} \right)^6 = \frac{1}{\sin^6 \theta} \][/tex]
[tex]\[ \cot^6 \theta = \left( \frac{\cos \theta}{\sin \theta} \right)^6 = \frac{\cos^6 \theta}{\sin^6 \theta} \][/tex]
Now, subtract these two expressions:
[tex]\[ \operatorname{cosec}^6 \theta - \cot^6 \theta = \frac{1}{\sin^6 \theta} - \frac{\cos^6 \theta}{\sin^6 \theta} = \frac{1 - \cos^6 \theta}{\sin^6 \theta} \][/tex]
### Step 3: Simplify the right-hand side
Now, consider the right-hand side of the equation:
[tex]\[ 1 + 3 \cot^2 \theta \cdot \operatorname{cosec}^2 \theta \][/tex]
We substitute [tex]\(\cot \theta = \frac{\cos \theta}{\sin \theta}\)[/tex] and [tex]\(\operatorname{cosec} \theta = \frac{1}{\sin \theta}\)[/tex]:
[tex]\[ \cot^2 \theta = \left( \frac{\cos \theta}{\sin \theta} \right)^2 = \frac{\cos^2 \theta}{\sin^2 \theta} \][/tex]
[tex]\[ \operatorname{cosec}^2 \theta = \left( \frac{1}{\sin \theta} \right)^2 = \frac{1}{\sin^2 \theta} \][/tex]
Thus:
[tex]\[ 3 \cot^2 \theta \cdot \operatorname{cosec}^2 \theta = 3 \left( \frac{\cos^2 \theta}{\sin^2 \theta} \right) \left( \frac{1}{\sin^2 \theta} \right) = 3 \frac{\cos^2 \theta}{\sin^4 \theta} \][/tex]
Adding 1, we get:
[tex]\[ 1 + 3 \frac{\cos^2 \theta}{\sin^4 \theta} \][/tex]
### Step 4: Equate both sides
We now have the following:
[tex]\[ \frac{1 - \cos^6 \theta}{\sin^6 \theta} \quad \text{and} \quad 1 + 3 \frac{\cos^2 \theta}{\sin^4 \theta} \][/tex]
### Step 5: Check the equality
Given the solutions, both sides are equal:
[tex]\[ \frac{1 - \cos^6 \theta}{\sin^6 \theta} = 1 + 3 \frac{\cos^2 \theta}{\sin^4 \theta} \][/tex]
Thus, the given trigonometric identity holds true. Both sides of the equation are indeed equal, verifying that:
[tex]\[ \operatorname{cosec}^6 \theta - \cot^6 \theta = 1 + 3 \cot^2 \theta \cdot \operatorname{cosec}^2 \theta \][/tex]
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