The value of $$\frac{\cos^260^{\circ}+4\sec^230^{\circ}-\tan^245^{\circ}}{\sin^230^{\circ}+\cos^230^{\circ}}$$

We need to find value of $$\frac{\cos^260^{\circ}+4\sec^230^{\circ}-\tan^245^{\circ}}{\sin^230^{\circ}+\cos^230^{\circ}}$$

we know that $$sin^2 \theta + cos^2 \theta$$ = 1

So, $$\frac{\cos^260^{\circ}+4\sec^230^{\circ}-\tan^245^{\circ}}{\sin^230^{\circ}+\cos^230^{\circ}}$$ = $$\frac{\cos^260^{\circ}+4\sec^230^{\circ}-\tan^245^{\circ}}{1}$$

$$cos^2 60 = \frac{1}{2}^2$$

$$sec^2 30 = \frac{2}{\surd 3}$$

$$tan^2 45$$ = 1

So, $$\frac{\cos^260^{\circ}+4\sec^230^{\circ}-\tan^245^{\circ}}{1}$$ = $$\frac{1}{4} + 4\frac{4}{3} - 1^2$$

=$$\frac{55}{12}$$