Question

yuda por favoooooooooooor

Answer 1

Respuesta:

 $-csc^2x$

Explicación paso a paso:

$\cfrac{\sin^2x+\sin^2x\cdot\cos^2x+\cos^4x+\tan^2x}{\cos^2x+\sin^4x+\sin^2x\cdot\cos^2x-\sec^2x}\\\\\\\cfrac{\sin^2x+(\sin^2x\cdot\cos^2x+\cos^4x)+\tan^2x}{\cos^2x+(\sin^4x+\sin^2x\cdot\cos^2x)-\sec^2x}\\\\\\\cfrac{\sin^2x+[\cos^2x\cdot(\sin^2x+\cos^2x)]+\tan^2x}{\cos^2x+[\sin^2x\cdot(\sin^2x+\cos^2x)]-\sec^2x}$

$\cfrac{\sin^2x+[\cos^2x\cdot(1)]+\tan^2x}{\cos^2x+[\sin^2x\cdot(1)]-\sec^2x}\\\\\\\cfrac{\sin^2x+\cos^2x+\tan^2x}{\cos^2x+\sin^2x-\sec^2x}\\\\\\\cfrac{(\sin^2x+\cos^2x)+\tan^2x}{(\cos^2x+\sin^2x)-\sec^2x}\\\\\\\cfrac{1+\tan^2x}{1-\sec^2x}$

$\cfrac{1+\tan^2x}{1-\sec^2x}\\\\\\\cfrac{1+\cfrac{\sin^2x}{\cos^2x}}{1-\cfrac{1}{cos^2x}}\\\\\\\cfrac{~~\cfrac{\cos^2x+\sin^2x}{\cos^2x}~~}{\cfrac{\cos^2x-1}{cos^2x}}\\\\\\\cfrac{~\cos^2x+\sin^2x~}{\cos^2x-1}\\\\\\\cfrac{~1~}{\cos^2x-1}\\\\\\\cfrac{~1~}{-(1-\cos^2x)}\\\\\\\cfrac{~1~}{-(\sin^2x)}$

$-\cfrac{~1~}{\sin^2x}\\\\-\csc^2x$