<s>[INST] Limit of $\mathbf{tan}^2(\theta)[1-\mathbf{sin}(\theta)]$ Evaulate the limit of $\mathbf{tan}^2(\theta)[1-\mathbf{sin}(\theta)]$.

$$\lim \limits_{\theta \to \frac{\pi}{2}} \mathbf{tan}^2(\theta)[1-\mathbf{sin}(\theta)] $$

I have attempted the problem by direct substitution and get:
$$\mathbf{tan}^2(\frac{\pi}2)[1-\mathbf{sin}(\frac{\pi}2)] $$
$$\infty^2 \cdot [0]$$
$$\infty \cdot 0$$
I do not know whether to state that it is zero, but I graphed it and got $\frac{1}{2}$.  The limit has to be evaluated algebraically.
 [/INST] $$\lim _{ \theta \to \frac { \pi  }{ 2 }  }{ \tan } ^{ 2 }(\theta )[1-{ \sin }(\theta )]=\lim _{ \theta \to \frac { \pi  }{ 2 }  }{ \tan } ^{ 2 }(\theta )\frac { 1-\sin ^{ 2 }{ (\theta ) }  }{ 1+{ \sin }(\theta ) } =\lim _{ \theta \to \frac { \pi  }{ 2 }  }{ \tan } ^{ 2 }(\theta )\frac { \cos ^{ 2 }{ (\theta ) }  }{ 1+{ \sin }(\theta ) } =\\ =\lim _{ \theta \to \frac { \pi  }{ 2 }  }{ \frac { \sin ^{ 2 }{ (\theta ) }  }{ 1+{ \sin }(\theta ) }  } =\frac { 1 }{ 2 } $$
Another way is to apply L'hospital's rule 
$$\lim _{ \theta \to \frac { \pi  }{ 2 }  }{ \tan } ^{ 2 }(\theta )\left[ 1-\sin { \left( \theta  \right)  }  \right] =\lim _{ \theta \to \frac { \pi  }{ 2 }  }{ \frac { 1-\sin { \left( \theta  \right)  }  }{ \cot ^{ 2 }{ \left( \theta  \right)  }  }  } =\lim _{ \theta \to \frac { \pi  }{ 2 }  }{ \frac { -\cos { \left( \theta  \right)  }  }{ 2\cot { \left( \theta  \right) \left( -\frac { 1 }{ \sin ^{ 2 }{ \left( \theta  \right)  }  }  \right)  }  }  } =\frac { 1 }{ 2 } $$
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