Question

Select the correct statement(s) for the equation $1-2x-x^2={\tan }^2(x+y)+{\cot }^2(x+y)?$

• A. $y=-1+n\pi \pm \frac{\pi }{4},n\in Z$
• B. $y=1+n\pi \pm \frac{\pi }{4},n\in Z$
• C. Exactly one value of $x$ exists
• D. Exactly two value of $x$ exists

Answer 1

Answer: (B), (C)

Exactly one value of $x$ exists

Given that $1-2x-x^2={\tan }^2(x+y)+{\cot }^2(x+y)$
Using $a^2+b^2=(a-b{)}^2+2ab$
$\Rightarrow 1-2x-x^2=\left[\tan \right(x+y)-\cot (x+y){]}^2+2\tan (x+y\left)\cot \right(x+y)$
$\Rightarrow 1-2x-x^2=\left[\tan \right(x+y)-\cot (x+y){]}^2+2$
$\Rightarrow -2+1-2x-x^2=\left[\tan \right(x+y)-\cot (x+y){]}^2$
$\Rightarrow -1-2x-x^2=\left[\tan \right(x+y)-\cot (x+y){]}^2$
$\Rightarrow -(x+1{)}^2=[\tan (x+y)-\cot (x+y){]}^2$
$\Rightarrow (x+1{)}^2+[\tan (x+y)-\cot (x+y){]}^2=0$
$\Rightarrow (x+1{)}^2=0\&[\tan (x+y)-\cot (x+y){]}^2=0$
$\Rightarrow x=-1\&\tan (x+y)=\cot (x+y)$
$\Rightarrow {\tan }^2(x+y)=1$
${\tan }^2(x+y)={\tan }^2\frac{\pi }{4}$
$\Rightarrow x+y=n\pi \pm \frac{\pi }{4},n\in Z$
$\Rightarrow -1+y=n\pi \pm \frac{\pi }{4},n\in Z$
$\Rightarrow y=1+n\pi \pm \frac{\pi }{4},n\in Z$​​​​​