Question

1508 The equation of straight line with gradient 1 , passing through $\left(\frac{ m }{2}, \frac{ n }{2}\right)$ where $m , n \in R$ satisfies the equation $\sec ^2(n(m+2))+m^2=1\left(\right.$ where $\left.n \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\right)$ can be

• A. $x+y=0$
• B. $x-y=0$
• C. $x-y+\frac{\pi}{4}=0$
• D. $x-y-\frac{\pi}{4}=0$

Answer 1

Answer: (B), (C), (D)

$ \tan ^2(m+2) n+m^2=0 $
$m=0, \tan ^2(m+2) n=0$
$\Rightarrow \tan ^2 2 n=0 $
$\Rightarrow n=0, \frac{\pi}{2},-\frac{\pi}{2} $
$(m, n)=(0,0)(0, \pi / 2),(0,-\pi / 2) $
$\text { Equation of straight lines: } $
$y-0=1(x-0) \Rightarrow x-y=0$
$y-\frac{\pi}{4}=1(x-0) \Rightarrow x-y+\frac{\pi}{4}=0 $
$y+\frac{\pi}{4}=1(x-0) \Rightarrow x-y-\frac{\pi}{4}=0$