Can I get help on this? **Verify the Identity**Given the trigonometric identity:\[ ( \tan \theta + \cot \theta )^2 = \sec^2 \theta + \csc^2 \theta \]To verify this identity, we will manipulate both sides of the equation and show that they are indeed equal. **Left-Hand Side (LHS) Manipulation:**\[ ( \tan \theta + \cot \theta )^2 \]Expand the squared term:\[ = \tan^2 \theta + 2 \tan \theta \cot \theta + \cot^2 \theta \]Since \( \cot \theta = \frac{1}{\tan \theta} \), substitute \( \cot \theta \) in the expression:\[ = \tan^2 \theta + 2 \tan \theta \cdot \frac{1}{\tan \theta} + \left( \frac{1}{\tan \theta} \right)^2 \]\[ = \tan^2 \theta + 2 + \frac{1}{\tan^2 \theta} \]\[ = \tan^2 \theta + 2 + \cot^2 \theta \]**Right-Hand Side (RHS) Manipulation:**\[ \sec^2 \theta + \csc^2 \theta \]We know the basic trigonometric identities:\[ \sec^2 \theta = 1 + \tan^2 \theta \]\[ \csc^2 \theta = 1 + \cot^2 \theta \]Substitute \( \sec^2 \theta \) and \( \csc^2 \theta \) in the expression:\[ 1 + \tan^2 \theta + 1 + \cot^2 \theta \] \[ = \tan^2 \theta + \cot^2 \theta + 2 \]**Conclusion:**Both the LHS and the RHS simplify to the same expression:\[ \tan^2 \theta + cot^2 \theta + 2 \]Hence, the trigonometric identity is verified:\[ ( \tan \theta + \cot \theta )^2 = \sec^2 \theta + \csc^2 \theta \]

Name: Can I get help on this? **Verify the Identity**Given the trigonometric
Uploaded: 2024-03-20T07:06:27.000Z
Channel: Bartleby
Description: Can I get help on this? **Verify the Identity**Given the trigonometric identity:\[ ( \tan \theta + \cot \theta )^2 = \sec^2 \theta + \csc^2 \theta \]To verify this identity, we will manipulate both sides of the equation and show that they are indeed