The value of: $$\frac{\sin 44^\circ}{\cos 46^\circ} + \sin^2 60^\circ - \cos^2 45^\circ + \sec 60^\circ$$ is equal to:

Given that,

$$\frac{\sin 44^\circ}{\cos 46^\circ} + \sin^2 60^\circ - \cos^2 45^\circ + \sec 60^\circ$$

$$\sin(90^\circ-\theta)=\cos(\theta)$$

$$\Rightarrow \frac{\cos (90-46)^\circ}{\cos 46^\circ} + \sin^2 60^\circ - \cos^2 45^\circ + \sec 60^\circ$$

$$\Rightarrow \frac{\cos 46^\circ}{\cos 46^\circ} + \sin^2 60^\circ - \cos^2 45^\circ + \sec 60^\circ$$

$$\Rightarrow 1 + \sin^2 60^\circ - \cos^2 45^\circ + \sec 60^\circ$$

$$\Rightarrow 1 +\dfrac{3}{4} - \dfrac{1}{2} +2$$

$$\Rightarrow \dfrac{4+3-2+8}{4}=\dfrac{13}{4}$$