Question

For the equation $1-2x-x^2={\tan }^2(x+y)+{\cot }^2(x+y)$

• A. exactly one value of $x$ exists
• B. exactly two values of $x$ exists
• C. $y=-1+n\pi +\frac{\pi }{4},nϵZ$
• D. $y=1+n\pi +\frac{\pi }{4},nϵZ$

Answer 1

Answer: (B), (C)

exactly two values of $x$ exists
$y=-1+n\pi +\frac{\pi }{4},nϵZ$

The given equation can be written as
$\Rightarrow 3-2x-x^2=1+{\tan }^2(x+y)+1+{\cot }^2(x+y)\Rightarrow 4-{(x+1)}^2={\sec }^2(x+y)+{\csc }^2(x+y)\Rightarrow {\sin }^2(x+y){\cos }^2(x+y)\{4-{(x+1)}^2\}=1$
$\Rightarrow {\sin }^2(2x+2y)\{4-{(x+1)}^2\}=4$ ...(1)
Since ${\sin }^2(2x+2y)\le 1$ and $\{4-{(x+1)}^2\}\le 4$
Therefore (1) holds only when ${\sin }^2(2x+2y)=1$ ...(2)
and $\{4-{(x+1)}^2\}=4$ ...(3)
From (3), we get $x=-1$
Putting this in (2), we get $\sin (2y-2)=\pm 1$
$\Rightarrow 2y-2=n\pi \pm \frac{\pi }{2},n\in Z$
$\Rightarrow y=1+(2n\pm 1)\frac{\pi }{4},n\in Z$
$\therefore x=-1,y=1+(2n\pm 1)\frac{\pi }{4},n\in Z$