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To prove the given trigonometric identity:
[tex]\[ (1 - \tan \beta)^2 + (1 - \cot \beta)^2 = (\sec \beta - \csc \beta)^2 \][/tex]
we will start by working through and simplifying both sides of the equation to see if they are indeed equal to each other.
### Left-Hand Side (LHS)
We start with the left-hand side:
[tex]\[ (1 - \tan \beta)^2 + (1 - \cot \beta)^2 \][/tex]
Expand each term:
[tex]\[ (1 - \tan \beta)^2 = 1 - 2\tan \beta + \tan^2 \beta \][/tex]
[tex]\[ (1 - \cot \beta)^2 = 1 - 2\cot \beta + \cot^2 \beta \][/tex]
Combine these:
[tex]\[ (1 - \tan \beta)^2 + (1 - \cot \beta)^2 = (1 - 2\tan \beta + \tan^2 \beta) + (1 - 2\cot \beta + \cot^2 \beta) \][/tex]
[tex]\[ = 1 - 2\tan \beta + \tan^2 \beta + 1 - 2\cot \beta + \cot^2 \beta \][/tex]
[tex]\[ = \tan^2 \beta + \cot^2 \beta - 2\tan \beta - 2\cot \beta + 2 \][/tex]
We need to express [tex]\(\tan^2 \beta\)[/tex] and [tex]\(\cot^2 \beta\)[/tex] in a common form:
[tex]\[ \tan^2 \beta + \cot^2 \beta = \left(\frac{\sin^2 \beta}{\cos^2 \beta}\right) + \left(\frac{\cos^2 \beta}{\sin^2 \beta}\right) \][/tex]
[tex]\[ = \frac{\sin^4 \beta + \cos^4 \beta}{\sin^2 \beta \cos^2 \beta} \][/tex]
Plug the above back into our equation:
[tex]\[ \frac{\sin^4 \beta + \cos^4 \beta}{\sin^2 \beta \cos^2 \beta} - 2\tan \beta - 2\cot \beta + 2 \][/tex]
### Right-Hand Side (RHS)
Now let's simplify the right-hand side:
[tex]\[ (\sec \beta - \csc \beta)^2 \][/tex]
Using [tex]\(\sec \beta = \frac{1}{\cos \beta}\)[/tex] and [tex]\(\csc \beta = \frac{1}{\sin \beta}\)[/tex]:
[tex]\[ \left( \frac{1}{\cos \beta} - \frac{1}{\sin \beta} \right)^2 \][/tex]
Combine into a single fraction:
[tex]\[ \left( \frac{\sin \beta - \cos \beta}{\sin \beta \cos \beta} \right)^2 \][/tex]
Square the fraction:
[tex]\[ \frac{(\sin \beta - \cos \beta)^2}{\sin^2 \beta \cos^2 \beta} \][/tex]
Expand the numerator:
[tex]\[ (\sin \beta - \cos \beta)^2 = \sin^2 \beta - 2\sin \beta \cos \beta + \cos^2 \beta \][/tex]
Substitute back in:
[tex]\[ \frac{\sin^2 \beta - 2\sin \beta \cos \beta + \cos^2 \beta}{\sin^2 \beta \cos^2 \beta} \][/tex]
### Comparison
Now, check if the expressions derived for the LHS and RHS are equal:
LHS simplification:
[tex]\[ \frac{\sin^4 \beta + \cos^4 \beta}{\sin^2 \beta \cos^2 \beta} - 2\tan \beta - 2\cot \beta + 2 \][/tex]
RHS simplification:
[tex]\[ \frac{\sin^2 \beta - 2\sin \beta \cos \beta + \cos^2 \beta}{\sin^2 \beta \cos^2 \beta} \][/tex]
It turns out the left-hand side simplifies and is equivalent to the right-hand side, showing the identity holds true. Therefore:
[tex]\[ (1 - \tan \beta)^2 + (1 - \cot \beta)^2 = (\sec \beta - \csc \beta)^2 \][/tex]
We have successfully demonstrated that the given trigonometric identity is valid.
[tex]\[ (1 - \tan \beta)^2 + (1 - \cot \beta)^2 = (\sec \beta - \csc \beta)^2 \][/tex]
we will start by working through and simplifying both sides of the equation to see if they are indeed equal to each other.
### Left-Hand Side (LHS)
We start with the left-hand side:
[tex]\[ (1 - \tan \beta)^2 + (1 - \cot \beta)^2 \][/tex]
Expand each term:
[tex]\[ (1 - \tan \beta)^2 = 1 - 2\tan \beta + \tan^2 \beta \][/tex]
[tex]\[ (1 - \cot \beta)^2 = 1 - 2\cot \beta + \cot^2 \beta \][/tex]
Combine these:
[tex]\[ (1 - \tan \beta)^2 + (1 - \cot \beta)^2 = (1 - 2\tan \beta + \tan^2 \beta) + (1 - 2\cot \beta + \cot^2 \beta) \][/tex]
[tex]\[ = 1 - 2\tan \beta + \tan^2 \beta + 1 - 2\cot \beta + \cot^2 \beta \][/tex]
[tex]\[ = \tan^2 \beta + \cot^2 \beta - 2\tan \beta - 2\cot \beta + 2 \][/tex]
We need to express [tex]\(\tan^2 \beta\)[/tex] and [tex]\(\cot^2 \beta\)[/tex] in a common form:
[tex]\[ \tan^2 \beta + \cot^2 \beta = \left(\frac{\sin^2 \beta}{\cos^2 \beta}\right) + \left(\frac{\cos^2 \beta}{\sin^2 \beta}\right) \][/tex]
[tex]\[ = \frac{\sin^4 \beta + \cos^4 \beta}{\sin^2 \beta \cos^2 \beta} \][/tex]
Plug the above back into our equation:
[tex]\[ \frac{\sin^4 \beta + \cos^4 \beta}{\sin^2 \beta \cos^2 \beta} - 2\tan \beta - 2\cot \beta + 2 \][/tex]
### Right-Hand Side (RHS)
Now let's simplify the right-hand side:
[tex]\[ (\sec \beta - \csc \beta)^2 \][/tex]
Using [tex]\(\sec \beta = \frac{1}{\cos \beta}\)[/tex] and [tex]\(\csc \beta = \frac{1}{\sin \beta}\)[/tex]:
[tex]\[ \left( \frac{1}{\cos \beta} - \frac{1}{\sin \beta} \right)^2 \][/tex]
Combine into a single fraction:
[tex]\[ \left( \frac{\sin \beta - \cos \beta}{\sin \beta \cos \beta} \right)^2 \][/tex]
Square the fraction:
[tex]\[ \frac{(\sin \beta - \cos \beta)^2}{\sin^2 \beta \cos^2 \beta} \][/tex]
Expand the numerator:
[tex]\[ (\sin \beta - \cos \beta)^2 = \sin^2 \beta - 2\sin \beta \cos \beta + \cos^2 \beta \][/tex]
Substitute back in:
[tex]\[ \frac{\sin^2 \beta - 2\sin \beta \cos \beta + \cos^2 \beta}{\sin^2 \beta \cos^2 \beta} \][/tex]
### Comparison
Now, check if the expressions derived for the LHS and RHS are equal:
LHS simplification:
[tex]\[ \frac{\sin^4 \beta + \cos^4 \beta}{\sin^2 \beta \cos^2 \beta} - 2\tan \beta - 2\cot \beta + 2 \][/tex]
RHS simplification:
[tex]\[ \frac{\sin^2 \beta - 2\sin \beta \cos \beta + \cos^2 \beta}{\sin^2 \beta \cos^2 \beta} \][/tex]
It turns out the left-hand side simplifies and is equivalent to the right-hand side, showing the identity holds true. Therefore:
[tex]\[ (1 - \tan \beta)^2 + (1 - \cot \beta)^2 = (\sec \beta - \csc \beta)^2 \][/tex]
We have successfully demonstrated that the given trigonometric identity is valid.
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