mplipy as for as $$ \frac{ an ^{2}(360+ heta)}{\cos (90+ heta) \cdot \sin 11} $$ $$ \frac{2 \cos [-150]+ an }{\sin ^{2} 135-\cos } $$

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mplipy as for as $$ \frac{	an ^{2}(360+	heta)}{\cos (90+	heta) \cdot \sin 11} $$ $$ \frac{2 \cos [-150]+	an }{\sin ^{2} 135-\cos } $$

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**Step 1. Simplify the first expression**

We start with
$$
\frac{\tan ^{2}(360+\theta)}{\cos (90+\theta) \cdot \sin 11}.
$$

1.1 Because the tangent function has a period of $$180^\circ$$, we have 
$$
\tan(360+\theta)=\tan\theta.
$$
Thus,
$$
\tan^2(360+\theta)=\tan^2\theta.
$$

1.2 For the cosine in the denominator, use the identity
$$
\cos(90+\theta)=-\sin\theta.
$$
Then the denominator becomes
$$
(-\sin\theta)\cdot \sin 11.
$$

1.3 Write $$\tan^2\theta$$ in terms of sine and cosine:
$$
\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}.
$$

1.4 Now substitute back:
$$
\frac{\tan ^{2}(360+\theta)}{\cos (90+\theta) \cdot \sin 11}
=\frac{\frac{\sin^2\theta}{\cos^2\theta}}{-\sin\theta\,\sin 11}
=\frac{\sin^2\theta}{\cos^2\theta}\cdot\frac{-1}{\sin\theta\,\sin 11}.
$$

1.5 Cancel a factor of $$\sin\theta$$:
$$
=\frac{-\sin\theta}{\cos^2\theta\,\sin 11}.
$$

So the simplified form is
$$
\boxed{\frac{-\sin\theta}{\cos^2\theta\,\sin 11}}.
$$

---

**Step 2. Simplify the second expression**

We interpret the given expression as
$$
\frac{2\cos(-150)+\tan(-150)}{\sin^2(135)-\cos(-150)}.
$$

2.1 Evaluate $$\cos(-150)$$:  
Since cosine is even,
$$
\cos(-150)=\cos150.
$$
And noting that
$$
\cos150=-\cos30=-\frac{\sqrt{3}}{2},
$$
we obtain
$$
\cos(-150)=-\frac{\sqrt{3}}{2}.
$$

2.2 Evaluate $$\tan(-150)$$:  
Since tangent is odd,
$$
\tan(-150)=-\tan150.
$$
Now, using
$$
\tan150=\tan(180-30)=-\tan30=-\frac{1}{\sqrt{3}},
$$
we have
$$
\tan(-150)= -\left(-\frac{1}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}.
$$

2.3 Evaluate the numerator:
$$
2\cos(-150)+\tan(-150)=2\left(-\frac{\sqrt{3}}{2}\right)+\frac{1}{\sqrt{3}}
=-\sqrt{3}+\frac{1}{\sqrt{3}}.
$$
Writing $$-\sqrt{3}$$ as $$-\frac{3}{\sqrt{3}}$$, we combine:
$$
-\frac{3}{\sqrt{3}}+\frac{1}{\sqrt{3}}=-\frac{2}{\sqrt{3}}.
$$

2.4 Evaluate the denominator:

First, compute $$\sin135$$. Since
$$
\sin135=\sin(180-45)=\sin45=\frac{\sqrt{2}}{2},
$$
then
$$
\sin^2(135)=\left(\frac{\sqrt{2}}{2}\right)^2=\frac{1}{2}.
$$
Also, we already have
$$
\cos(-150)=-\frac{\sqrt{3}}{2}.
$$
Thus,
$$
\sin^2(135)-\cos(-150)=\frac{1}{2}-\left(-\frac{\sqrt{3}}{2}\right)
=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}.
$$

2.5 Combine numerator and denominator:
$$
\frac{-\frac{2}{\sqrt{3}}}{\frac{1+\sqrt{3}}{2}}
=-\frac{2}{\sqrt{3}}\cdot\frac{2}{1+\sqrt{3}}
=-\frac{4}{\sqrt{3}(1+\sqrt{3})}.
$$

2.6 To rationalize, multiply numerator and denominator by $$1-\sqrt{3}$$:
$$
-\frac{4}{\sqrt{3}(1+\sqrt{3})}\cdot\frac{1-\sqrt{3}}{1-\sqrt{3}}
=-\frac{4(1-\sqrt{3})}{\sqrt{3}(1-3)}
=-\frac{4(1-\sqrt{3})}{\sqrt{3}(-2)}
=\frac{4(1-\sqrt{3})}{2\sqrt{3}}
=\frac{2(1-\sqrt{3})}{\sqrt{3}}.
$$

Thus, the second expression simplifies to
$$
\boxed{\frac{2(1-\sqrt{3})}{\sqrt{3}}}.
$$
The simplified forms are: 1. $$ \frac{-\sin\theta}{\cos^2\theta\,\sin 11} $$ 2. $$ \frac{2(1-\sqrt{3})}{\sqrt{3}} $$

mplipy as for as \( \frac{\tan ^{2}(360+\theta)}{\cos (90+\theta) \cdot \sin 11} \) \( \frac{2 \cos [-150]+\tan }{\sin ^{2} 135-\cos } \)

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