Trigonometry & Inverse Trigonometry Question - None

Question

Statement I: If $a, b, c \in \mathbb{R}$ and not all equal, then $\frac{bc + ca + ab}{a^2 + b^2 + c^2} < 1$

Statement II: $\sec \theta < -1$ and $\sec \theta > 1$

Accepted Answer

Solution: We know that $a^2 + b^2 + c^2 - ab - bc - ca \geq 0$ for all real $a, b, c$, with equality only when $a = b = c$.

Since $a, b, c$ are not all equal:

$a^2 + b^2 + c^2 - ab - bc - ca > 0$

$\Rightarrow a^2 + b^2 + c^2 > ab + bc + ca$

$\Rightarrow \frac{bc + ca + ab}{a^2 + b^2 + c^2} < 1$

Statement I is true. Statement II is true (the range of secant is $(-\infty, -1] \cup [1, \infty)$). Both statements are true and Statement II correctly explains Statement I. The answer is D.

Question:

Step-by-Step Solution

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