Let $P(\alpha, \beta)$ and $Q(\gamma, \delta)$ be two points that lie on the curve $\tan ^2(x+y)+\cos…
Let $P(\alpha, \beta)$ and $Q(\gamma, \delta)$ be two points that lie on the curve $\tan ^2(x+y)+\cos ^2(x+y)$ $+y^2+2 y=0$ in the $X Y$-plane. If the distance between $P$ and $Q$ is $d$, then $\cos d=$
0
$(-1)^n, n \in N$
$\pm \pi$
$\pm 2 n \pi, n \in N$
Solution
We have, $\tan ^2(x+y)+\cos ^2(x+y)+y^2+2 y=0$
$
\begin{aligned}
& \Rightarrow \quad \sec ^2(x+y)-1+\cos ^2(x+y)+y^2+2 y=0 \\
& \Rightarrow \quad \sec ^2(x+y)+\cos ^2(x+y)+y^2+2 y+1=2
\end{aligned}
$
Now, as minimum value of $\sec ^2(x+y)+\cos ^2(x+y)$ is 2.
$
\begin{aligned}
& \therefore \quad x+y=0 \text { and } y^2+2 y+1=0 \\
& \Rightarrow \quad x=-y \text { and }(y+1)^2=0 \\
& \Rightarrow \quad x=-y \quad \text { and } y=-1 \\
&
\end{aligned}
$
Thus, $x=1$ and $y=-1$
Hence, the points $P$ and $Q$ coincides, and so $d=0$
Now, again if we take $x+Y=\pi$ and $y^2+2 y+1=0$, then $x=\pi+1$ and $y=-1$ Similarly, if $x+y=2 \pi$ and $y^2+2 y+1=0$ Then, $x=2 \pi+1, y=-1$ $\therefore \quad d=\pi$
