The equation of straight lines passing through ordered pairs $(a,b)$ satisfying equation $\sec^2((a+1)b) + a^2 - 1 = 0$, and having slope $\frac{1}{2}$, is (are):
Step-by-Step Solution
Key Concept: A sum of squares equals zero only when each square term individually equals zero, constraining both parameters.
From $\sec^2(a+1)b + a^2 - 1 = 0$, rewrite as $1 + \tan^2(a+1)b + a^2 - 1 = 0$, giving $\tan^2(a+1)b + a^2 = 0$. This requires $a = 0$ and $\tan(a+1)b = 0$, so $a = 0$ and $b = n\pi$ for integer $n$. Lines pass through points $(0, n\pi)$ with slope $\frac{1}{2}$, giving equations $y - n\pi = \frac{1}{2}(x - 0)$ or $x - 2y + 2n\pi = 0$. For $n = 0, 1, -1$, the lines are $x - 2y = 0$, $x - 2y + 2\pi = 0$, and $x - 2y - 2\pi = 0$ respectively.
Correct Answer: 1,3,4