Join IDNLearn.com today and start getting the answers you've been searching for. Join our community to receive prompt and reliable responses to your questions from experienced professionals.
Sagot :
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.
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Find the answers you need at IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.
