The value of the following is:
$$\frac{(\tan 20^\circ)^2}{(\cosec 70^\circ)^2} + \frac{(\cot 20^\circ)^2}{(\sec 70^\circ)^2} + 2 \tan 15^\circ. \tan 45^\circ. \tan 75^\circ$$

$$\frac{(\tan 20^\circ)^2}{(\cosec 70^\circ)^2} + \frac{(\cot 20^\circ)^2}{(\sec 70^\circ)^2} + 2 \tan 15^\circ. \tan 45^\circ. \tan 75^\circ$$

$$=\frac{\tan^2\left(90-70\right)^{\circ}}{\operatorname{cosec}^270^{\circ}}+\frac{\cot^2\left(90-70\right)^{\circ}}{\sec^270^{\circ}}+2\tan15^{\circ}.\tan45^{\circ}.\tan\left(90-15\right)^{\circ}$$

$$=\frac{\cot^270^{\circ}}{\operatorname{cosec}^270^{\circ}}+\frac{\tan^270^{\circ}}{\sec^270^{\circ}}+2\tan15^{\circ}.\tan45^{\circ}.\cot15^{\circ}$$

$$=\frac{\operatorname{cosec}^270^{\circ}-1}{\operatorname{cosec}^270^{\circ}}+\frac{\sec^270^{\circ}-1}{\sec^270^{\circ}}+2\tan45^{\circ}$$

$$=1-\frac{1}{\operatorname{cosec}^270^{\circ}}+1-\frac{1}{\sec^270^{\circ}}+2\left(1\right)$$

$$=4-\left(\sin^270^{\circ}+\cos^270^{\circ}\right)$$

$$=4-1$$

$$=3$$