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 ${\sec }^2(n(m+2))+m^2=1($ where $n\in \left[\frac{-\pi }{2},\frac{\pi }{2}\right])$ 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)

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