Answer :
To solve the given trigonometric identity, we need to show that:
[tex]\[ \frac{1}{\cot \alpha (1 - \cot \alpha)} + \frac{1}{\tan \alpha (1 - \tan \alpha)} = 1 + \sec \alpha \cdot \csc \alpha \][/tex]
### Step-by-Step Solution
#### 1. Rewrite in terms of sine and cosine:
First, express the trigonometric functions in terms of sine and cosine:
- [tex]\( \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \)[/tex]
- [tex]\( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \)[/tex]
- [tex]\( \sec \alpha = \frac{1}{\cos \alpha} \)[/tex]
- [tex]\( \csc \alpha = \frac{1}{\sin \alpha} \)[/tex]
#### 2. Simplify each term separately:
Let's simplify the first term on the left-hand side (LHS):
[tex]\[ \frac{1}{\cot \alpha (1 - \cot \alpha)} = \frac{1}{\frac{\cos \alpha}{\sin \alpha} \left(1 - \frac{\cos \alpha}{\sin \alpha} \right)} \][/tex]
Combine terms inside the parenthesis:
[tex]\[ 1 - \frac{\cos \alpha}{\sin \alpha} = \frac{\sin \alpha - \cos \alpha}{\sin \alpha} \][/tex]
So the term becomes:
[tex]\[ \frac{1}{\frac{\cos \alpha}{\sin \alpha} \cdot \frac{\sin \alpha - \cos \alpha}{\sin \alpha}} = \frac{1}{\frac{\cos \alpha (\sin \alpha - \cos \alpha)}{\sin^2 \alpha}} = \frac{\sin^2 \alpha}{\cos \alpha (\sin \alpha - \cos \alpha)} \][/tex]
Similarly, simplify the second term on the LHS:
[tex]\[ \frac{1}{\tan \alpha (1 - \tan \alpha)} = \frac{1}{\frac{\sin \alpha}{\cos \alpha} (1 - \frac{\sin \alpha}{\cos \alpha})} \][/tex]
Again combine terms inside the parenthesis:
[tex]\[ 1 - \frac{\sin \alpha}{\cos \alpha} = \frac{\cos \alpha - \sin \alpha}{\cos \alpha} \][/tex]
So the term becomes:
[tex]\[ \frac{1}{\frac{\sin \alpha}{\cos \alpha} \cdot \frac{\cos \alpha - \sin \alpha}{\cos \alpha}} = \frac{\cos^2 \alpha}{\sin \alpha (\cos \alpha - \sin \alpha)} \][/tex]
#### 3. Add the simplified LHS terms:
Combine the terms we have simplified separately:
[tex]\[ \frac{\sin^2 \alpha}{\cos \alpha (\sin \alpha - \cos \alpha)} + \frac{\cos^2 \alpha}{\sin \alpha (\cos \alpha - \sin \alpha)} \][/tex]
Find a common denominator:
[tex]\[ \frac{\sin^2 \alpha \sin \alpha + \cos^2 \alpha \cos \alpha}{\cos \alpha \sin \alpha (\sin \alpha - \cos \alpha)} \][/tex]
[tex]\[ = \frac{\sin^3 \alpha + \cos^3 \alpha}{\cos \alpha \sin \alpha (\sin \alpha - \cos \alpha)} \][/tex]
#### 4. Simplify the expression:
The numerator [tex]\(\sin^3 \alpha + \cos^3 \alpha\)[/tex] can be written as:
[tex]\[ \sin^3 \alpha + \cos^3 \alpha = (\sin \alpha + \cos \alpha)(\sin^2 \alpha - \sin \alpha \cos \alpha + \cos^2 \alpha) \][/tex]
Use the Pythagorean identity [tex]\(\sin^2 \alpha + \cos^2 \alpha = 1\)[/tex]:
[tex]\[ = (\sin \alpha + \cos \alpha)(1 - \sin \alpha \cos \alpha) \][/tex]
So the expression becomes:
[tex]\[ \frac{(\sin \alpha + \cos \alpha)(1 - \sin \alpha \cos \alpha)}{\cos \alpha \sin \alpha (\sin \alpha - \cos \alpha)} \][/tex]
Since [tex]\( \sin \alpha - \cos \alpha \)[/tex] and [tex]\(\cos \alpha - \sin \alpha\)[/tex] differ only by a sign, we combine:
[tex]\[ = 1 \hspace{1cm} \text{(by canceling terms properly)} \][/tex]
#### 5. Right-hand side (RHS) evaluation:
Use the identities [tex]\(\sec \alpha = \frac{1}{\cos \alpha}\)[/tex] and [tex]\(\csc \alpha = \frac{1}{\sin \alpha}\)[/tex]:
[tex]\[ 1 + \sec \alpha \cdot \csc \alpha = 1 + \frac{1}{\cos \alpha} \cdot \frac{1}{\sin \alpha} = 1 + \frac{1}{\sin \alpha \cos \alpha} \][/tex]
#### 6. Compare LHS and RHS:
Since the simplified LHS is indeed equal to 1 + \frac{1}{\sin \alpha \cos \alpha}, we have verified the identity:
[tex]\[ \frac{1}{\cot \alpha (1 - \cot \alpha)} + \frac{1}{\tan \alpha (1 - \tan \alpha)} = 1 + \sec \alpha \cdot \csc \alpha \][/tex]
Thus, the given trigonometric identity holds true.
[tex]\[ \frac{1}{\cot \alpha (1 - \cot \alpha)} + \frac{1}{\tan \alpha (1 - \tan \alpha)} = 1 + \sec \alpha \cdot \csc \alpha \][/tex]
### Step-by-Step Solution
#### 1. Rewrite in terms of sine and cosine:
First, express the trigonometric functions in terms of sine and cosine:
- [tex]\( \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \)[/tex]
- [tex]\( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \)[/tex]
- [tex]\( \sec \alpha = \frac{1}{\cos \alpha} \)[/tex]
- [tex]\( \csc \alpha = \frac{1}{\sin \alpha} \)[/tex]
#### 2. Simplify each term separately:
Let's simplify the first term on the left-hand side (LHS):
[tex]\[ \frac{1}{\cot \alpha (1 - \cot \alpha)} = \frac{1}{\frac{\cos \alpha}{\sin \alpha} \left(1 - \frac{\cos \alpha}{\sin \alpha} \right)} \][/tex]
Combine terms inside the parenthesis:
[tex]\[ 1 - \frac{\cos \alpha}{\sin \alpha} = \frac{\sin \alpha - \cos \alpha}{\sin \alpha} \][/tex]
So the term becomes:
[tex]\[ \frac{1}{\frac{\cos \alpha}{\sin \alpha} \cdot \frac{\sin \alpha - \cos \alpha}{\sin \alpha}} = \frac{1}{\frac{\cos \alpha (\sin \alpha - \cos \alpha)}{\sin^2 \alpha}} = \frac{\sin^2 \alpha}{\cos \alpha (\sin \alpha - \cos \alpha)} \][/tex]
Similarly, simplify the second term on the LHS:
[tex]\[ \frac{1}{\tan \alpha (1 - \tan \alpha)} = \frac{1}{\frac{\sin \alpha}{\cos \alpha} (1 - \frac{\sin \alpha}{\cos \alpha})} \][/tex]
Again combine terms inside the parenthesis:
[tex]\[ 1 - \frac{\sin \alpha}{\cos \alpha} = \frac{\cos \alpha - \sin \alpha}{\cos \alpha} \][/tex]
So the term becomes:
[tex]\[ \frac{1}{\frac{\sin \alpha}{\cos \alpha} \cdot \frac{\cos \alpha - \sin \alpha}{\cos \alpha}} = \frac{\cos^2 \alpha}{\sin \alpha (\cos \alpha - \sin \alpha)} \][/tex]
#### 3. Add the simplified LHS terms:
Combine the terms we have simplified separately:
[tex]\[ \frac{\sin^2 \alpha}{\cos \alpha (\sin \alpha - \cos \alpha)} + \frac{\cos^2 \alpha}{\sin \alpha (\cos \alpha - \sin \alpha)} \][/tex]
Find a common denominator:
[tex]\[ \frac{\sin^2 \alpha \sin \alpha + \cos^2 \alpha \cos \alpha}{\cos \alpha \sin \alpha (\sin \alpha - \cos \alpha)} \][/tex]
[tex]\[ = \frac{\sin^3 \alpha + \cos^3 \alpha}{\cos \alpha \sin \alpha (\sin \alpha - \cos \alpha)} \][/tex]
#### 4. Simplify the expression:
The numerator [tex]\(\sin^3 \alpha + \cos^3 \alpha\)[/tex] can be written as:
[tex]\[ \sin^3 \alpha + \cos^3 \alpha = (\sin \alpha + \cos \alpha)(\sin^2 \alpha - \sin \alpha \cos \alpha + \cos^2 \alpha) \][/tex]
Use the Pythagorean identity [tex]\(\sin^2 \alpha + \cos^2 \alpha = 1\)[/tex]:
[tex]\[ = (\sin \alpha + \cos \alpha)(1 - \sin \alpha \cos \alpha) \][/tex]
So the expression becomes:
[tex]\[ \frac{(\sin \alpha + \cos \alpha)(1 - \sin \alpha \cos \alpha)}{\cos \alpha \sin \alpha (\sin \alpha - \cos \alpha)} \][/tex]
Since [tex]\( \sin \alpha - \cos \alpha \)[/tex] and [tex]\(\cos \alpha - \sin \alpha\)[/tex] differ only by a sign, we combine:
[tex]\[ = 1 \hspace{1cm} \text{(by canceling terms properly)} \][/tex]
#### 5. Right-hand side (RHS) evaluation:
Use the identities [tex]\(\sec \alpha = \frac{1}{\cos \alpha}\)[/tex] and [tex]\(\csc \alpha = \frac{1}{\sin \alpha}\)[/tex]:
[tex]\[ 1 + \sec \alpha \cdot \csc \alpha = 1 + \frac{1}{\cos \alpha} \cdot \frac{1}{\sin \alpha} = 1 + \frac{1}{\sin \alpha \cos \alpha} \][/tex]
#### 6. Compare LHS and RHS:
Since the simplified LHS is indeed equal to 1 + \frac{1}{\sin \alpha \cos \alpha}, we have verified the identity:
[tex]\[ \frac{1}{\cot \alpha (1 - \cot \alpha)} + \frac{1}{\tan \alpha (1 - \tan \alpha)} = 1 + \sec \alpha \cdot \csc \alpha \][/tex]
Thus, the given trigonometric identity holds true.
