The value of $$\sec^2 28^\circ - \cot^2 62^\circ + \sin^2 60^\circ + \cosec^2 30^\circ$$ is equal to:

$$\sec^2 28^\circ - \cot^2 62^\circ + \sin^2 60^\circ + \cosec^2 30^\circ$$

We know that $$\sec(90^\circ -A)=\cosec A, \cosec(90^\circ -A)=\sec A$$ and $$\sin^2 A+\cos^2 A=1 $$

Now, $$\sec^2 28^\circ - \cot^2 62^\circ + \sin^2 60^\circ + \cosec^2 30^\circ$$

$$\Rightarrow \cosec^2 (90^\circ-28^\circ) - \cot^2 62^\circ + \sin^2 60^\circ + \sec^2 (90^\circ- 30^\circ$$

$$\Rightarrow \cosec^2 62^\circ - \cot^2 62^\circ + \sin^2 60^\circ + \sec^2 60^\circ$$

$$\Rightarrow \dfrac{1}{\sin^2 62^\circ}-\dfrac{\cos^2 62^\circ}{\sin^2 62^\circ} + \sin^2 60^\circ + \dfrac{1}{\cos^2 60^\circ}$$

$$\Rightarrow \dfrac{1-\cos^2 62^\circ}{\sin^2 62^\circ}+ \sin^2 60^\circ + \dfrac{1}{\cos^2 60}$$

$$\Rightarrow \dfrac{\sin^2 62^\circ}{\sin^2 62^\circ}+ \sin^2 60^\circ + \dfrac{1}{\cos^2 60}$$

$$\Rightarrow 1+(\dfrac{\sqrt 3}{2})^2 + (\dfrac{1}{\dfrac{1}{2}})^2$$

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

$$\Rightarrow \dfrac{4+3+16}{4}$$

$$\Rightarrow \dfrac{23}{4}$$