Como simplificar \( \sin^2\theta(1+\tan^2\theta)\) usando identidades trigonométricas?

Usando \(1+\tan^2\theta=\sec^2\theta=\dfrac{1}{\cos^2\theta}\), obtém-se \( \sin^2\theta(1+\tan^2\theta)=\dfrac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta\).

Explicação e exemplo prático:
Aplicando a identidade, substituímos \(1+\tan^2\theta\) por \(\dfrac{1}{\cos^2\theta}\), assim
\( \sin^2\theta(1+\tan^2\theta)=\sin^2\theta\cdot\dfrac{1}{\cos^2\theta}=\dfrac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta\).

Exemplo:
Considere \(\theta=30^\circ\), então \(\sin\theta=\tfrac{1}{2}\), \(\cos\theta=\tfrac{\sqrt{3}}{2}\), \(\tan\theta=\tfrac{1}{\sqrt{3}}\).
Calculando, \( \sin^2\theta(1+\tan^2\theta)=\left(\tfrac{1}{2}\right)^2\!\left(1+\left(\tfrac{1}{\sqrt{3}}\right)^2\right)=\tfrac{1}{4}\left(1+\tfrac{1}{3}\right)=\tfrac{1}{4}\cdot\tfrac{4}{3}=\tfrac{1}{3}\),
e \(\tan^2\theta=\left(\tfrac{1}{\sqrt{3}}\right)^2=\tfrac{1}{3}\), confirma-se a igualdade.