JEE Advanced Physics: Trigonometric Functions; Limits questions with solutions

Q1. If tan²(alpha) + cot²(alpha) equals the limit as x approaches infinity of (sqrt(x² + 8x + 3) - sqrt(x² + 4x + 2)), where alpha belongs to (-pi/2, pi/2), then the value of (1/2)(tan(alpha) + cot(alpha)) is:

1. 0
2. 1
3. 4
4. 7

Answer: 1

Limit: multiply by conjugate. (sqrt(x²+8x+3) - sqrt(x²+4x+2)) = (4x+1) / (sqrt(x²+8x+3) + sqrt(x²+4x+2)). As x → infinity, this → 4x / 2x = 2. So tan²(alpha) + cot²(alpha) = 2. The minimum value of tan²(alpha) + cot²(alpha) is 2 (by AM-GM), achieved only when tan(alpha) = cot(alpha) = 1, i.e., alpha = pi/4. Then (1/2)(tan(alpha) + cot(alpha)) = (1/2)(1 + 1) = 1.