Answer :
To prove the identity [tex]\(\tan^2 \theta - \cot^2 \theta = \sec^2 \theta (1 - \cot^2 \theta)\)[/tex], let's proceed with the following steps:
Step 1: Define the functions involved in the identity.
- [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex]
- [tex]\(\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}\)[/tex]
- [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex]
Step 2: Express the left-hand side (LHS) of the identity.
The LHS is given by:
[tex]\[ \tan^2 \theta - \cot^2 \theta \][/tex]
Step 3: Rewrite [tex]\(\tan^2 \theta\)[/tex] and [tex]\(\cot^2 \theta\)[/tex]:
[tex]\[ \tan^2 \theta = \left(\frac{\sin \theta}{\cos \theta}\right)^2 = \frac{\sin^2 \theta}{\cos^2 \theta} \][/tex]
[tex]\[ \cot^2 \theta = \left(\frac{\cos \theta}{\sin \theta}\right)^2 = \frac{\cos^2 \theta}{\sin^2 \theta} \][/tex]
Putting these together in the LHS expression gives:
[tex]\[ \tan^2 \theta - \cot^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta} \][/tex]
Step 4: Express the right-hand side (RHS) of the identity.
The RHS is given by:
[tex]\[ \sec^2 \theta (1 - \cot^2 \theta) \][/tex]
Step 5: Rewrite [tex]\(\sec^2 \theta\)[/tex]:
[tex]\[ \sec^2 \theta = \left(\frac{1}{\cos \theta}\right)^2 = \frac{1}{\cos^2 \theta} \][/tex]
And we already know [tex]\(\cot^2 \theta\)[/tex]:
[tex]\[ \cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta} \][/tex]
Step 6: Combining these in the RHS expression gives:
[tex]\[ \sec^2 \theta (1 - \cot^2 \theta) = \frac{1}{\cos^2 \theta} \left(1 - \frac{\cos^2 \theta}{\sin^2 \theta}\right) \][/tex]
Simplify the expression inside the parentheses first:
[tex]\[ 1 - \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{\sin^2 \theta - \cos^2 \theta}{\sin^2 \theta} \][/tex]
Now, substituting back, we get:
[tex]\[ \sec^2 \theta (1 - \cot^2 \theta) = \frac{1}{\cos^2 \theta} \cdot \frac{\sin^2 \theta - \cos^2 \theta}{\sin^2 \theta} = \frac{\sin^2 \theta - \cos^2 \theta}{\cos^2 \theta \sin^2 \theta} \][/tex]
Step 7: Simplify the RHS expression:
[tex]\[ \frac{\sin^2 \theta - \cos^2 \theta}{\cos^2 \theta \sin^2 \theta} \][/tex]
Step 8: Revisit and simplify the LHS for comparison:
LHS:
[tex]\[ \frac{\sin^2 \theta}{\cos^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{\sin^4 \theta - \cos^4 \theta}{\cos^2 \theta \sin^2 \theta} \][/tex]
By factoring identities:
[tex]\[ \sin^4 \theta - \cos^4 \theta = (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta) \][/tex]
Using [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex]:
[tex]\[ \sin^4 \theta - \cos^4 \theta = (\sin^2 \theta - \cos^2 \theta) \][/tex]
Thus:
[tex]\[ \tan^2 \theta - \cot^2 \theta = \frac{\sin^2 \theta - \cos^2 \theta}{\cos^2 \theta \sin^2 \theta} \][/tex]
Step 9: Verify if LHS equals RHS.
Both expressions for LHS and RHS simplify to:
[tex]\[ \frac{\sin^2 \theta - \cos^2 \theta}{\cos^2 \theta \sin^2 \theta} \][/tex]
Since they are equal, the identity is proven:
[tex]\[ \tan^2 \theta - \cot^2 \theta = \sec^2 \theta (1 - \cot^2 \theta) \][/tex]
Step 1: Define the functions involved in the identity.
- [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex]
- [tex]\(\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}\)[/tex]
- [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex]
Step 2: Express the left-hand side (LHS) of the identity.
The LHS is given by:
[tex]\[ \tan^2 \theta - \cot^2 \theta \][/tex]
Step 3: Rewrite [tex]\(\tan^2 \theta\)[/tex] and [tex]\(\cot^2 \theta\)[/tex]:
[tex]\[ \tan^2 \theta = \left(\frac{\sin \theta}{\cos \theta}\right)^2 = \frac{\sin^2 \theta}{\cos^2 \theta} \][/tex]
[tex]\[ \cot^2 \theta = \left(\frac{\cos \theta}{\sin \theta}\right)^2 = \frac{\cos^2 \theta}{\sin^2 \theta} \][/tex]
Putting these together in the LHS expression gives:
[tex]\[ \tan^2 \theta - \cot^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta} \][/tex]
Step 4: Express the right-hand side (RHS) of the identity.
The RHS is given by:
[tex]\[ \sec^2 \theta (1 - \cot^2 \theta) \][/tex]
Step 5: Rewrite [tex]\(\sec^2 \theta\)[/tex]:
[tex]\[ \sec^2 \theta = \left(\frac{1}{\cos \theta}\right)^2 = \frac{1}{\cos^2 \theta} \][/tex]
And we already know [tex]\(\cot^2 \theta\)[/tex]:
[tex]\[ \cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta} \][/tex]
Step 6: Combining these in the RHS expression gives:
[tex]\[ \sec^2 \theta (1 - \cot^2 \theta) = \frac{1}{\cos^2 \theta} \left(1 - \frac{\cos^2 \theta}{\sin^2 \theta}\right) \][/tex]
Simplify the expression inside the parentheses first:
[tex]\[ 1 - \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{\sin^2 \theta - \cos^2 \theta}{\sin^2 \theta} \][/tex]
Now, substituting back, we get:
[tex]\[ \sec^2 \theta (1 - \cot^2 \theta) = \frac{1}{\cos^2 \theta} \cdot \frac{\sin^2 \theta - \cos^2 \theta}{\sin^2 \theta} = \frac{\sin^2 \theta - \cos^2 \theta}{\cos^2 \theta \sin^2 \theta} \][/tex]
Step 7: Simplify the RHS expression:
[tex]\[ \frac{\sin^2 \theta - \cos^2 \theta}{\cos^2 \theta \sin^2 \theta} \][/tex]
Step 8: Revisit and simplify the LHS for comparison:
LHS:
[tex]\[ \frac{\sin^2 \theta}{\cos^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{\sin^4 \theta - \cos^4 \theta}{\cos^2 \theta \sin^2 \theta} \][/tex]
By factoring identities:
[tex]\[ \sin^4 \theta - \cos^4 \theta = (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta) \][/tex]
Using [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex]:
[tex]\[ \sin^4 \theta - \cos^4 \theta = (\sin^2 \theta - \cos^2 \theta) \][/tex]
Thus:
[tex]\[ \tan^2 \theta - \cot^2 \theta = \frac{\sin^2 \theta - \cos^2 \theta}{\cos^2 \theta \sin^2 \theta} \][/tex]
Step 9: Verify if LHS equals RHS.
Both expressions for LHS and RHS simplify to:
[tex]\[ \frac{\sin^2 \theta - \cos^2 \theta}{\cos^2 \theta \sin^2 \theta} \][/tex]
Since they are equal, the identity is proven:
[tex]\[ \tan^2 \theta - \cot^2 \theta = \sec^2 \theta (1 - \cot^2 \theta) \][/tex]
