Question

If $acosx+cotx+1=cosecx$ then the possible values of x can be

• A. $x=n\pi ,a\epsilon R,n\epsilon |$
• B. $x=(4n+1)\frac{\pi }{2},a\epsilon R,n\epsilon |$
• C. $x=\frac{1}{2}si{n}^{-1}\left(\frac{4(a+1)}{a^2}\right),a\epsilon (-\infty ,2-2\sqrt{2})\cup [2+2\sqrt{2},\infty ]-\{-1\}$
• D. none of these

Answer 1

Answer: (B), (C)

$x=\frac{1}{2}si{n}^{-1}\left(\frac{4(a+1)}{a^2}\right),a\epsilon (-\infty ,2-2\sqrt{2})\cup [2+2\sqrt{2},\infty ]-\{-1\}$
$x=(4n+1)\frac{\pi }{2},a\epsilon R,n\epsilon |$

$a\cos x+\cot x+1=\csc x$
$\Rightarrow a\cos x+\frac{\cos x}{\sin x}+1=\frac{1}{\sin x}$
$\Rightarrow a\cos x\sin x+\cos x+\sin x=1\Rightarrow \sin x+\cos x=1-a\cos x\sin x$
$\Rightarrow {(\sin x+\cos x)}^2={(1-a\cos x\sin x)}^2\Rightarrow 1+2\sin x\cos x=1+a^2{\cos }^{2}x{\sin }^{2}x-2a\sin x\cos x$
$\Rightarrow a^{2}si{n}^{2}xco{s}^2-2(a+1)sinxcosx=0\Rightarrow sin2x\left(\frac{a^2}{2}sin2x-2(a+1)\right)=0$
$sin2x=0$ $\forall$ $a$ $ϵR$
$\Rightarrow x=\frac{n\pi }{2},nϵI$
And $sin2x=\frac{4(a+1)}{a^2}\Rightarrow aϵ(-\infty ,2-2\sqrt{2},\infty )$