Respuesta :
[tex]\bf cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}
\qquad\qquad
sec(\theta)=\cfrac{1}{cos(\theta)}\quad\qquad 1+tan^2(\theta)=sec^2(\theta)\\\\\\
tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}
\qquad \qquad cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}
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cot(x)sec^4(x)=cot(x)+2tan(x)+tan^3(x)\\\\
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[tex]\bf \textit{so, let's do the left-hand-side}\\\\ cot(x)sec^2(x)sec^2(x)\implies cot(x)[1+tan^2(x)][1+tan^2(x)] \\\\\\ cot(x)[1^2+2tan^2(x)+tan^4(x)] \\\\\\ cot(x)+2tan^2(x)cot(x)+tan^4(x)cot(x) \\\\\\ cot(x)+2\cdot \cfrac{sin^2(x)}{cos^2(x)}\cdot \cfrac{cos(x)}{sin(x)}+\cfrac{sin^4(x)}{cos^4(x)}\cdot \cfrac{cos(x)}{sin(x)} \\\\\\ cot(x)+2\cdot \cfrac{sin(x)}{cos(x)}+\cfrac{sin^3(x)}{cos^3(x)}\implies \boxed{cot(x)+2tan(x)+tan^3(x)}[/tex]
[tex]\bf \textit{so, let's do the left-hand-side}\\\\ cot(x)sec^2(x)sec^2(x)\implies cot(x)[1+tan^2(x)][1+tan^2(x)] \\\\\\ cot(x)[1^2+2tan^2(x)+tan^4(x)] \\\\\\ cot(x)+2tan^2(x)cot(x)+tan^4(x)cot(x) \\\\\\ cot(x)+2\cdot \cfrac{sin^2(x)}{cos^2(x)}\cdot \cfrac{cos(x)}{sin(x)}+\cfrac{sin^4(x)}{cos^4(x)}\cdot \cfrac{cos(x)}{sin(x)} \\\\\\ cot(x)+2\cdot \cfrac{sin(x)}{cos(x)}+\cfrac{sin^3(x)}{cos^3(x)}\implies \boxed{cot(x)+2tan(x)+tan^3(x)}[/tex]
