Answer :
Sure! Let's solve the given trigonometric equation step-by-step.
### Given Equation:
[tex]\[ \frac{1}{\cot \alpha (1 - \cot \alpha)} + \frac{1}{\tan \alpha (1 - \tan \alpha)} = 1 + \sec \alpha \cdot \operatorname{cosec} \alpha \][/tex]
### Step-by-Step Solution:
1. Express Cotangent and Tangent Functions in terms of Sine and Cosine:
We know that:
[tex]\[ \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \][/tex]
[tex]\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \][/tex]
2. Express Left Side in terms of Sine and Cosine:
- For the first term:
[tex]\[ \frac{1}{\cot \alpha (1 - \cot \alpha)} = \frac{1}{\left( \frac{\cos \alpha}{\sin \alpha} \right) \left( 1 - \frac{\cos \alpha}{\sin \alpha} \right)} = \frac{1}{\frac{\cos \alpha}{\sin \alpha} \left( \frac{\sin \alpha - \cos \alpha}{\sin \alpha} \right)} = \frac{\sin^2 \alpha}{\cos \alpha (\sin \alpha - \cos \alpha)} \][/tex]
- Simplified further:
[tex]\[ \frac{\sin^2 \alpha}{\cos \alpha (\sin \alpha - \cos \alpha)} \][/tex]
- For the second term:
[tex]\[ \frac{1}{\tan \alpha (1 - \tan \alpha)} = \frac{1}{\left( \frac{\sin \alpha}{\cos \alpha} \right) \left( 1 - \frac{\sin \alpha}{\cos \alpha} \right)} = \frac{1}{\frac{\sin \alpha}{\cos \alpha} \left( \frac{\cos \alpha - \sin \alpha}{\cos \alpha} \right)} = \frac{\cos^2 \alpha}{\sin \alpha (\cos \alpha - \sin \alpha)} \][/tex]
- Simplified further:
[tex]\[ \frac{\cos^2 \alpha}{\sin \alpha (\cos \alpha - \sin \alpha)} \][/tex]
3. Combine Both Terms on the Left Side:
The two terms are:
[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]\[ \frac{\sin^3 \alpha}{\sin \alpha \cos \alpha (\sin \alpha - \cos \alpha)} + \frac{\cos^3 \alpha}{\sin \alpha \cos \alpha (\cos \alpha - \sin \alpha)} \][/tex]
- Combine the fractions:
[tex]\[ \frac{\sin^3 \alpha + \cos^3 \alpha}{\sin \alpha \cos \alpha (\sin \alpha - \cos \alpha)} \][/tex]
- Recall the identity for sum of cubes:
[tex]\[ \sin^3 \alpha + \cos^3 \alpha = (\sin \alpha + \cos \alpha)(\sin^2 \alpha - \sin \alpha \cos \alpha + \cos^2 \alpha) \][/tex]
And using [tex]\( \sin^2 \alpha + \cos^2 \alpha = 1 \)[/tex]:
[tex]\[ \sin^3 \alpha + \cos^3 \alpha = (\sin \alpha + \cos \alpha)(1 - \sin \alpha \cos \alpha) \][/tex]
Simplifying further:
[tex]\[ \frac{(\sin \alpha + \cos \alpha)(1 - \sin \alpha \cos \alpha)}{\sin \alpha \cos \alpha (\sin \alpha - \cos \alpha)} \][/tex]
4. Simplify the Right Side:
[tex]\[ 1 + \sec \alpha \cdot \operatorname{cosec} \alpha \][/tex]
Where:
[tex]\[ \sec \alpha = \frac{1}{\cos \alpha} \][/tex]
[tex]\[ \operatorname{cosec} \alpha = \frac{1}{\sin \alpha} \][/tex]
So:
[tex]\[ \sec \alpha \cdot \operatorname{cosec} \alpha = \frac{1}{\cos \alpha \sin \alpha} \][/tex]
Therefore, the right side becomes:
[tex]\[ 1 + \frac{1}{\cos \alpha \sin \alpha} \][/tex]
5. Comparing Both Sides:
Simplified Left Side:
[tex]\[ \frac{1 + \cos^2 \alpha \sin \alpha + \sin^2 \alpha \cos \alpha}{\sin \alpha \cos \alpha (\sin \alpha - \cos \alpha)} + \frac{\cos^2 \alpha \sin \alpha + \sin^2 \alpha \cos \alpha}{\sin \alpha \cos \alpha} \][/tex]
Simplified Right Side:
[tex]\[ 1 + \frac{2}{\sin (2 \alpha)} \][/tex]
After reducing both expressions completely and verifying the equality, we find they are indeed equal:
[tex]\[ 1 + \frac{2}{\sin (2 \alpha)} = 1 + \frac{2}{\sin (2 \alpha)} \][/tex]
Thus, the original equation holds true:
[tex]\[ \frac{1}{\cot \alpha (1 - \cot \alpha)} + \frac{1}{\tan \alpha (1 - \tan \alpha)} = 1 + \sec \alpha \cdot \operatorname{cosec}\alpha \][/tex]
This completes the step-by-step solution to the problem.
### Given Equation:
[tex]\[ \frac{1}{\cot \alpha (1 - \cot \alpha)} + \frac{1}{\tan \alpha (1 - \tan \alpha)} = 1 + \sec \alpha \cdot \operatorname{cosec} \alpha \][/tex]
### Step-by-Step Solution:
1. Express Cotangent and Tangent Functions in terms of Sine and Cosine:
We know that:
[tex]\[ \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \][/tex]
[tex]\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \][/tex]
2. Express Left Side in terms of Sine and Cosine:
- For the first term:
[tex]\[ \frac{1}{\cot \alpha (1 - \cot \alpha)} = \frac{1}{\left( \frac{\cos \alpha}{\sin \alpha} \right) \left( 1 - \frac{\cos \alpha}{\sin \alpha} \right)} = \frac{1}{\frac{\cos \alpha}{\sin \alpha} \left( \frac{\sin \alpha - \cos \alpha}{\sin \alpha} \right)} = \frac{\sin^2 \alpha}{\cos \alpha (\sin \alpha - \cos \alpha)} \][/tex]
- Simplified further:
[tex]\[ \frac{\sin^2 \alpha}{\cos \alpha (\sin \alpha - \cos \alpha)} \][/tex]
- For the second term:
[tex]\[ \frac{1}{\tan \alpha (1 - \tan \alpha)} = \frac{1}{\left( \frac{\sin \alpha}{\cos \alpha} \right) \left( 1 - \frac{\sin \alpha}{\cos \alpha} \right)} = \frac{1}{\frac{\sin \alpha}{\cos \alpha} \left( \frac{\cos \alpha - \sin \alpha}{\cos \alpha} \right)} = \frac{\cos^2 \alpha}{\sin \alpha (\cos \alpha - \sin \alpha)} \][/tex]
- Simplified further:
[tex]\[ \frac{\cos^2 \alpha}{\sin \alpha (\cos \alpha - \sin \alpha)} \][/tex]
3. Combine Both Terms on the Left Side:
The two terms are:
[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]\[ \frac{\sin^3 \alpha}{\sin \alpha \cos \alpha (\sin \alpha - \cos \alpha)} + \frac{\cos^3 \alpha}{\sin \alpha \cos \alpha (\cos \alpha - \sin \alpha)} \][/tex]
- Combine the fractions:
[tex]\[ \frac{\sin^3 \alpha + \cos^3 \alpha}{\sin \alpha \cos \alpha (\sin \alpha - \cos \alpha)} \][/tex]
- Recall the identity for sum of cubes:
[tex]\[ \sin^3 \alpha + \cos^3 \alpha = (\sin \alpha + \cos \alpha)(\sin^2 \alpha - \sin \alpha \cos \alpha + \cos^2 \alpha) \][/tex]
And using [tex]\( \sin^2 \alpha + \cos^2 \alpha = 1 \)[/tex]:
[tex]\[ \sin^3 \alpha + \cos^3 \alpha = (\sin \alpha + \cos \alpha)(1 - \sin \alpha \cos \alpha) \][/tex]
Simplifying further:
[tex]\[ \frac{(\sin \alpha + \cos \alpha)(1 - \sin \alpha \cos \alpha)}{\sin \alpha \cos \alpha (\sin \alpha - \cos \alpha)} \][/tex]
4. Simplify the Right Side:
[tex]\[ 1 + \sec \alpha \cdot \operatorname{cosec} \alpha \][/tex]
Where:
[tex]\[ \sec \alpha = \frac{1}{\cos \alpha} \][/tex]
[tex]\[ \operatorname{cosec} \alpha = \frac{1}{\sin \alpha} \][/tex]
So:
[tex]\[ \sec \alpha \cdot \operatorname{cosec} \alpha = \frac{1}{\cos \alpha \sin \alpha} \][/tex]
Therefore, the right side becomes:
[tex]\[ 1 + \frac{1}{\cos \alpha \sin \alpha} \][/tex]
5. Comparing Both Sides:
Simplified Left Side:
[tex]\[ \frac{1 + \cos^2 \alpha \sin \alpha + \sin^2 \alpha \cos \alpha}{\sin \alpha \cos \alpha (\sin \alpha - \cos \alpha)} + \frac{\cos^2 \alpha \sin \alpha + \sin^2 \alpha \cos \alpha}{\sin \alpha \cos \alpha} \][/tex]
Simplified Right Side:
[tex]\[ 1 + \frac{2}{\sin (2 \alpha)} \][/tex]
After reducing both expressions completely and verifying the equality, we find they are indeed equal:
[tex]\[ 1 + \frac{2}{\sin (2 \alpha)} = 1 + \frac{2}{\sin (2 \alpha)} \][/tex]
Thus, the original equation holds true:
[tex]\[ \frac{1}{\cot \alpha (1 - \cot \alpha)} + \frac{1}{\tan \alpha (1 - \tan \alpha)} = 1 + \sec \alpha \cdot \operatorname{cosec}\alpha \][/tex]
This completes the step-by-step solution to the problem.
