Certainly! Let's dive into the given expression and show the step-by-step solution to verify the given trigonometric identity:
[tex]\[ \frac{\cot^3 B - \tan^3 B}{\cot B - \tan B} = \sec^2 B + \csc^2 B - 1 \][/tex]
### Step 1: Simplify the Left-Hand Side (LHS)
The left-hand side of the equation is:
[tex]\[ \frac{\cot^3 B - \tan^3 B}{\cot B - \tan B} \][/tex]
Notice that this fraction is in the form of a difference of cubes. Recall the algebraic identity for the difference of cubes:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
Here, let [tex]\( a = \cot B \)[/tex] and [tex]\( b = \tan B \)[/tex]. Applying the difference of cubes identity:
[tex]\[ \cot^3 B - \tan^3 B = (\cot B - \tan B)(\cot^2 B + \cot B \cdot \tan B + \tan^2 B) \][/tex]
Now, substitute this back into the fraction:
[tex]\[ \frac{(\cot B - \tan B)(\cot^2 B + \cot B \cdot \tan B + \tan^2 B)}{\cot B - \tan B} \][/tex]
The common factor [tex]\( (\cot B - \tan B) \)[/tex] cancels out:
[tex]\[ \cot^2 B + \cot B \cdot \tan B + \tan^2 B \][/tex]
### Step 2: Further Simplify Using Trigonometric Identities
We know that:
[tex]\[ \cot B = \frac{1}{\tan B} \][/tex]
So:
[tex]\[ \cot^2 B = \left( \frac{1}{\tan B} \right)^2 = \frac{1}{\tan^2 B} \][/tex]
Now, substitute these into the simplified LHS expression:
[tex]\[ \frac{1}{\tan^2 B} + \left( \frac{1}{\tan B} \cdot \tan B \right) + \tan^2 B \][/tex]
[tex]\[ \frac{1}{\tan^2 B} + 1 + \tan^2 B \][/tex]
We can rewrite this as:
[tex]\[ \tan^2 B + 1 + \frac{1}{\tan^2 B} \][/tex]
### Step 3: Simplify the Right-Hand Side (RHS)
The right-hand side is given as:
[tex]\[ \sec^2 B + \csc^2 B - 1 \][/tex]
Recall the Pythagorean identities:
[tex]\[ \sec^2 B = 1 + \tan^2 B \][/tex]
[tex]\[ \csc^2 B = 1 + \cot^2 B \][/tex]
So:
[tex]\[ \sec^2 B + \csc^2 B - 1 = (1 + \tan^2 B) + (1 + \cot^2 B) - 1 \][/tex]
[tex]\[ = \tan^2 B + \cot^2 B + 1 \][/tex]
Substituting back, we get:
[tex]\[ \tan^2 B + \frac{1}{\tan^2 B} + 1 \][/tex]
### Step 4: Verify the Equality
We have already simplified both sides and can now compare:
[tex]\[ \frac{\cot^3 B - \tan^3 B}{\cot B - \tan B} = \tan^2 B + 1 + \frac{1}{\tan^2 B} \][/tex]
[tex]\[ \sec^2 B + \csc^2 B - 1 = \tan^2 B + \frac{1}{\tan^2 B} + 1 \][/tex]
Both the simplified left-hand side and right-hand side are indeed equal, proving the given trigonometric identity:
[tex]\[ \frac{\cot^3 B - \tan^3 B}{\cot B - \tan B} = \sec^2 B + \csc^2 B - 1 \][/tex]
Therefore, the identity is verified to be true.
