Question

Answer the question, it's urgent with solution.. I will very thankful

Answer 1

Answer:

2 OR 3 ik think so if you know then uh ok

Answer 2

Answer:

Step-by-step explanation:

We have,

$\tt{tan^{-1}\left(5+2\,sin(x)-sin^2(x)\right)+cot^{-1}\left(1+5^{sec^2(y)}\right)=\dfrac{\pi}{2}}$

$\sf{\blue{We\,\,\,know\,,}}\\\sf{\green{tan^{-1}(\alpha)+cot^{-1}(\beta)=\dfrac{\pi}{2},}}\\\sf{\green{iff\,\,\,\alpha=\beta}}$

So,

$\sf{5+2\,sin(x)-sin^2(x)=1+5^{sec^2(y)}}$

On comparing,

$\sf{sec^2(y)=1\,\,\,\,and\,\,\,\,2\,sin(x)-sin^2(x)=1}$

$\sf{\implies\,sec^2(y)=1\,\,\,\,and\,\,\,\,sin^2(x)-2\,sin(x)+1=0}$

$\sf{\implies\,sec^2(y)=1\,\,\,\,and\,\,\,\,\{sin(x)-1\}^2=0}$

$\sf{\implies\,cos^2(y)=1\,\,\,\,and\,\,\,\,sin(x)=1}$

$\sf{\implies\,y=n\pi\,\,\,\,and\,\,\,\,x=(4m+1)\dfrac{\pi}{2}\,\,\,\,\,\,\,\,\,\,,\forall\,\,n\in\mathbb{Z}\,\,\,\,\&\,\,\,\,m\in\mathbb{Z^{+}}\cup\{0\}}$

For minimum positive value, n=0 ,and m=0

So,

$\sf{2[x+y]=2\left[0+\dfrac{\pi}{2}\right]=2\left[\dfrac{\pi}{2}\right]}$

$\sf{=2\times1=2}$