Trigonometry
Trigonometry
Allen Star Batch
Grade 11
Question:
If $\sin x + \cos x + \tan x + \cot y = 4$, where $x, y \in [0, \frac{\pi}{2}]$, then $\tan(\frac{y}{2})$ is a root of the equation:
$a^2 + 2a + 1 = 0$
$a^2 + 2a - 1 = 0$
$2a^2 - 2a - 1 = 0$
None of these
Step-by-Step Solution
Key Concept: By AM-GM inequality, $\sin x + \csc x \geq 2\sqrt{\sin x \cdot \csc x} = 2$ with equality only when both terms are equal.
For $\sin x + \csc x \geq 2$ and $\tan y + \cot y \geq 2$ with $x, y \in [0, \frac{\pi}{2}]$, equality holds only when $\sin x = 1$ and $\tan y = 1$. This gives $x = \frac{\pi}{2}$ and $y = \frac{\pi}{4}$. Therefore, $\tan(\frac{\pi}{2}) = \tan(\frac{\pi}{8}) = \sqrt{2} - 1$, and we verify that $\sqrt{2} - 1$ satisfies $a^2 + 2a - 1 = 0$.
Correct Answer: 2