¿ Cómo resuelvo este Ejercicio de demostración trigonométrica?

Si vas con igualdades no es necesario demostrar 'de izquierda a derecha' y viceversa, ya que directamente una vez que demostraste una de ellas, la otra es lo mismo pero recorriéndolo del final hacia el principio

$$\begin{align}&\bigg(1+tg(x)\bigg)^2+\bigg(1+cotg(x)\bigg)^2=\bigg(1+\frac{sen((x)}{\cos(x)}\bigg)^2+\bigg(1+\frac{\cos(x)}{sen(x)}\bigg)^2=\\&\bigg(\frac{\cos(x)+sen((x)}{\cos(x)}\bigg)^2+\bigg(\frac{sen(x)+\cos(x)}{sen(x)}\bigg)^2=\\&\frac{\cos^2(x)+2cos(x)sen(x)+sen^2(x)}{\cos^2(x)}+\frac{sen^2(x)+2sen(x)\cos(x)+\cos^2(x)}{sen^2(x)}=\\&\frac{1+2cos(x)sen(x)}{\cos^2(x)}+\frac{1+2sen(x)\cos(x)}{sen^2(x)}=\\&\frac{sen^2(x)(1+2cos(x)sen(x))+ \cos^2(x)(1+2sen(x)\cos(x))}{\cos^2(x)sen^2(x)}=\\&\frac{(sen^2(x)+\cos^2(x))(1+2cos(x)sen(x))}{\cos^2(x)sen^2(x)}=\\&\frac{(1)(1+2cos(x)sen(x))}{\cos^2(x)sen^2(x)}=\frac{1+2cos(x)sen(x)}{\cos^2(x)sen^2(x)} (*)\\&Veamos\ ahora\\&(sec(x)+cosec(x))^2=\bigg(\frac{1}{\cos(x)}+\frac{1}{sen(x)}\bigg)^2=\\&\frac{1}{\cos^2(x)}+\frac{2}{\cos(x)sen(x)}+\frac{1}{sen^2(x)}=\\&\frac{\cos(x)sen^3(x)+2cos^2(x)sen^2(x)+\cos^3(x)sen(x)}{\cos^2(x)\cos(x)sen(x)sen^2(x)}=\\&\frac{\cos(x)sen(x)(sen^2(x)+2cos(x)sen(x)+\cos^2(x))}{\cos^2(x)\cos(x)sen(x)sen^2(x)}=\\&\frac{sen^2(x)+2cos(x)sen(x)+\cos^2(x)}{\cos^2(x)sen^2(x)}=\\&\frac{1+2cos(x)sen(x)}{\cos^2(x)sen^2(x)}= (**)\\&Como\ (*)=(**) \text{ queda demostrado}\end{align}$$

Salu2