Question

If $a\le ta{n}^{-1}x+co{t}^{-1}x+si{n}^{-1}x\le b$, then

• A. a = 0, b = π
• B. a =1, b= π/2
• C. a=π/4, b= π/2
• D. a = 0, b = 2π

Answer 1

The correct option is A

a = 0, b = π

$ta{n}^{-1}x+co{t}^{-1}x+si{n}^{-1}x=\frac{\pi }{2}+si{n}^{-1}x$ ...(1)
Since, $-\frac{\pi }{2}\le si{n}^{-1}x\le \frac{\pi }{2}\Rightarrow 0\le \frac{\pi }{2}+si{n}^{-1}x\le \pi$
$\Rightarrow 0\le ta{n}^{-1}x+co{t}^{-1}x+si{n}^{-1}x\le \pi$ [From Eqq.(i)]
$\therefore a=0$ and $b=\pi$