Respuesta :

Okay, here we have this:

Considering the provided expression, we are going to prove the identity, so we obtain the following:

[tex]\frac{csc\theta(sin^2\theta+cos^2\theta tan\theta)}{sin\theta+cos\theta}=1[/tex][tex]\frac{\frac{1}{sin\vartheta}(sin^2\theta+cos^2\theta\frac{sin\theta}{cos\theta})}{sin\theta+cos\theta}=1[/tex][tex]\frac{\frac{1}{sin\vartheta}(sin^2\theta+cos\text{ }\theta sin\theta)}{sin\theta+cos\theta}=1[/tex][tex]\frac{(\frac{sin^2\theta}{sin\theta}+\frac{cos\text{ }\theta sin\theta}{sin\theta})}{sin\theta+cos\theta}=1[/tex][tex]\frac{(sin\text{ }\theta+cos\text{ }\theta)}{sin\theta+cos\theta}=1[/tex][tex]\frac{1}{1}=1[/tex][tex]1=1[/tex]