Question

If \( \sin t + \cos t = \sqrt{2} \), then \( \tan t + \cot t \) is equal to

• A. \( \frac{1}{2} \)
• B. \( 1 \)
• C. \( \frac{3}{2} \)
• D. \( \frac{5}{2} \)
• E. \( 2 \)

Hint:

Use the identity \( \tan t + \cot t = \frac{1}{\sin t \cos t} \) to simplify expressions.

Answer 1

Answer: (E)

Step 1: Square Both Sides \[ (\sin t + \cos t)^2 = (\sqrt{2})^2 \] \[ \sin^2 t + \cos^2 t + 2 \sin t \cos t = 2 \] Using \( \sin^2 t + \cos^2 t = 1 \), we get: \[ 1 + 2 \sin t \cos t = 2 \] \[ 2 \sin t \cos t = 1 \] \[ \sin t \cos t = \frac{1}{2} \] Step 2: Compute \( \tan t + \cot t \) \[ \tan t + \cot t = \frac{\sin t}{\cos t} + \frac{\cos t}{\sin t} \] Using the identity: \[ \tan t + \cot t = \frac{\sin^2 t + \cos^2 t}{\sin t \cos t} \] \[ = \frac{1}{\frac{1}{2}} = 2 \]
Final Answer: \[ \boxed{2} \]