The curve represented by the equation $\dfrac{x^{2}}{\sin \sqrt{2}-\sin \sqrt{3}}+\dfrac{y^{2}}{\cos \sqrt{2}-\cos \sqrt{3}}=1$ is

General form of an Ellipse:

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, where both $a$ and $b$ should be positive.

If:

$\rarr a>b$, then the foci lies on the $x$-axis.

$\rarr b>a$, then the foci lies on the $y$-axis.

---

We know that:

$\pi^{c}$(radians) = $180{\degree}$

$\Rightarrow 1^{c}=\dfrac{180}{22} \times 7$

$= 57.27{\degree}$

---

$\dfrac{x^2}{\sin \sqrt 2 - \sin \sqrt3}+\dfrac{y^2}{\cos \sqrt2 - \cos \sqrt3}=1$

Here, $\sqrt 2$ and $\sqrt 3$ are in radians. Let's convert them into degrees.

$\Rightarrow \dfrac{x^2}{\sin (1.414 \times 57.27) - \sin (1.732 \times 57.2)}+\dfrac{y^2}{\cos (1.414 \times 57.27) - \cos(1.73 \times 57.2) }=1$

$\dfrac{x^2}{\sin (80.99) - \sin (99.19) }+\dfrac{y^2}{\cos (80.99) - \cos (99.19)}=1$

---

As per the trigonometric quadrant rule, the value of $\sin (80.99)$ is positive [$\because 80.99$ lies in the first quadrant] and so is $\sin (99.19)$ [$\because 99.19$ lies in the second quadrant].

Also, $\cos (80.99)$ is positive [$\because 80.99$ lies in the first quadrant] and $\cos (99.19)$ is negative [$\because90.99$ lies in the second quadrant].

$\therefore (\cos (80.99) - \cos (99.19))$ is positive.

Note that the value of $\sin (80.99)$ is greater than that of $\sin (99.19)$. How?

The maximum value that $\sin \theta$ can take is $1$ when $\theta = 90{\degree}$. The angle $80.99$ is closer to $90$ in comparison to $99.19$. Hence, the former is greater than the latter and their difference will be positive.

As both $a$ and $b$ are positive, the given curve is an Ellipse.

Also, as $a$ is the difference of two values and $b$ is the sum of two values, we can say that $b > a$. Therefore, the foci of this ellipse lies on the $y$ axis.

Hence, answer is option (a).