Certainly! Let's solve the given expression step-by-step and verify the result:
Given expression:
[tex]\[ \frac{\cot^3 B - \tan^3 B}{\cot B - \tan B} \][/tex]
1. Factor the numerator:
The numerator is in the form of a difference of cubes, which can be factored using the identity:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
Let [tex]\( a = \cot B \)[/tex] and [tex]\( b = \tan B \)[/tex], then:
[tex]\[ \cot^3 B - \tan^3 B = (\cot B - \tan B)(\cot^2 B + \cot B \tan B + \tan^2 B) \][/tex]
2. Substitute back into the expression:
Substitute the factor form back into the expression:
[tex]\[ \frac{(\cot B - \tan B)(\cot^2 B + \cot B \tan B + \tan^2 B)}{\cot B - \tan B} \][/tex]
Since [tex]\(\cot B - \tan B\)[/tex] appears in both the numerator and the denominator, it gets cancelled:
[tex]\[ \cot^2 B + \cot B \tan B + \tan^2 B \][/tex]
3. Simplify the resulting expression:
To simplify [tex]\( \cot^2 B + \cot B \tan B + \tan^2 B \)[/tex]:
- We know that [tex]\(\cot B = \frac{1}{\tan B}\)[/tex].
- Therefore, [tex]\(\cot B \tan B = \frac{1}{\tan B} \cdot \tan B = 1\)[/tex].
The expression simplifies to:
[tex]\[ \cot^2 B + 1 + \tan^2 B \][/tex]
4. Use trigonometric identities:
Recall the trigonometric identities:
[tex]\[ \cot^2 B = \csc^2 B - 1 \][/tex]
[tex]\[ \tan^2 B = \sec^2 B - 1 \][/tex]
Substitute these identities into the simplified expression:
[tex]\[ (\csc^2 B - 1) + 1 + (\sec^2 B - 1) \][/tex]
Simplify by combining like terms:
[tex]\[ \csc^2 B - 1 + 1 + \sec^2 B - 1 \][/tex]
[tex]\[ \csc^2 B + \sec^2 B - 1 \][/tex]
Thus, the given expression simplifies to:
[tex]\[ \sec^2 B + \csc^2 B - 1 \][/tex]
Therefore, the original equation holds:
[tex]\[ \frac{\cot^3 B - \tan^3 B}{\cot B - \tan B} = \sec^2 B + \csc^2 B - 1 \][/tex]
