Question

If ${\tan }^2(x+y)+{\cot }^2(x+y)=1+2x-x^2,$ then the correct option(s) is/are

• A. $x=0$
• B. $x=1$
• C. $y=n\pi -1+\frac{\pi }{4},n\in Z$
• D. $y=n\pi -1-\frac{\pi }{4},n\in Z$

Answer 1

Answer: (B), (C), (D)

$y=n\pi -1-\frac{\pi }{4},n\in Z$

${\tan }^2(x+y)+{\cot }^2(x+y)=1+2x-x^2$
$\Rightarrow \left[\tan \right(x+y)-\cot (x+y){]}^2+2=1+2x-x^2$
$\Rightarrow \left[\tan \right(x+y)-\cot (x+y){]}^2=-(x^2-2x+1)$
$\Rightarrow \left[\tan \right(x+y)-\cot (x+y){]}^2=-(x-1{)}^2$

Now,
$\text{L.H.S.}\ge 0$ and $\text{R.H.S.}\le 0$
So, solution exist only when $\text{R.H.S.}=\text{L.H.S.}=0$

For $\text{R.H.S.}=0$
$-(x-1{)}^2=0\Rightarrow x=1$

For $\text{L.H.S.}=0$
$\left[\tan \right(x+y)-\cot (x+y){]}^2=0$
$\Rightarrow \tan (x+y)-\cot (x+y)=0$
$\Rightarrow \tan (x+y)=\cot (x+y)$
$\Rightarrow {\tan }^2(x+y)=1={\tan }^2\frac{\pi }{4}$
$\Rightarrow x+y=n\pi \pm \frac{\pi }{4}$
Here, $x=1$
$\Rightarrow y=(n\pi -1)\pm \frac{\pi }{4},n\in Z$