$$ \left. \begin{array} { l } { ( \frac { 2 } { 2 } ) ( \frac { \sqrt { 2 } } { 2 } ) } \ { \frac { \sin ( 180 + heta ) } { \sin ( 180 ^ { \circ } + heta ) \cos ( 180 ^ { \circ } - heta ) } - \frac { 6 } { 4 } = \frac { \frac { 1 } { 2 } } { \frac { 1 } { 2 } } = 1 } \ { \frac { \sin ( 90 ^ { \circ } - heta ) \cos ( - heta ) } { ( \sin heta ) ( - \cos heta ) } - \frac { \sin ( 90 ^ { \circ } - heta ) \cdot \cos } { 1 } } \end{array} ight. $$

Question

$$ \left. \begin{array} { l } { ( \frac { 2 } { 2 } ) ( \frac { \sqrt { 2 } } { 2 } ) } \ { \frac { \sin ( 180 + 	heta ) } { \sin ( 180 ^ { \circ } + 	heta ) \cos ( 180 ^ { \circ } - 	heta ) } - \frac { 6 } { 4 } = \frac { \frac { 1 } { 2 } } { \frac { 1 } { 2 } } = 1 } \ { \frac { \sin ( 90 ^ { \circ } - 	heta ) \cos ( - 	heta ) } { ( \sin 	heta ) ( - \cos 	heta ) } - \frac { \sin ( 90 ^ { \circ } - 	heta ) \cdot \cos } { 1 } } \end{array} ight. $$

UpStudy AI · Accepted Answer

1. **First Expression**

$$
\left(\frac{2}{2}\right)\left(\frac{\sqrt{2}}{2}\right)
$$

Since

$$
\frac{2}{2}=1,
$$

we obtain

$$
1\cdot\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}.
$$

2. **Second Expression**

$$
\frac{\sin(180^\circ+\theta)}{\sin(180^\circ+\theta)\cos(180^\circ-\theta)}-\frac{6}{4}.
$$

Recall the angle–addition formulas:

$$
\sin(180^\circ+\theta)=-\sin\theta,\quad \cos(180^\circ-\theta)=-\cos\theta.
$$

Thus the denominator is

$$
\sin(180^\circ+\theta)\cos(180^\circ-\theta)=(-\sin\theta)(-\cos\theta)=\sin\theta\cos\theta.
$$

So the fraction becomes

$$
\frac{-\sin\theta}{\sin\theta\cos\theta}=-\frac{1}{\cos\theta}\quad (\text{for } \sin\theta\neq0).
$$

Also,

$$
\frac{6}{4}=\frac{3}{2}.
$$

Thus the whole expression simplifies to

$$
-\frac{1}{\cos\theta}-\frac{3}{2}.
$$

In the statement it is also written that

$$
\frac{\frac{1}{2}}{\frac{1}{2}}=1,
$$

so if one wishes to set the above expression equal to 1 then one may write

$$
-\frac{1}{\cos\theta}-\frac{3}{2}=1.
$$

This equation is equivalent to

$$
-\frac{1}{\cos\theta} = 1+\frac{3}{2}=\frac{5}{2},
$$

so that

$$
\frac{1}{\cos\theta}=-\frac{5}{2}\quad\Longrightarrow\quad \cos\theta=-\frac{2}{5}.
$$

3. **Third Expression**

$$
\frac{\sin(90^\circ-\theta)\cos(-\theta)}{(\sin\theta)(-\cos\theta)}-\sin(90^\circ-\theta)\cdot\cos\theta.
$$

Notice that

$$
\sin(90^\circ-\theta)=\cos\theta \quad \text{and} \quad \cos(-\theta)=\cos\theta.
$$

Thus the numerator of the first term is

$$
\cos\theta\cos\theta=\cos^2\theta,
$$

and the denominator is

$$
\sin\theta\cdot(-\cos\theta)=-\sin\theta\cos\theta.
$$

Therefore, the first term becomes

$$
\frac{\cos^2\theta}{-\sin\theta\cos\theta}=-\frac{\cos\theta}{\sin\theta}=-\cot\theta.
$$

The second term is

$$
\sin(90^\circ-\theta)\cdot\cos\theta=\cos\theta\cdot\cos\theta=\cos^2\theta.
$$

Hence the whole expression simplifies to

$$
-\cot\theta-\cos^2\theta.
$$

---

Thus the simplified results are:

1. $$\displaystyle \frac{\sqrt{2}}{2}$$.

2. $$\displaystyle -\frac{1}{\cos\theta}-\frac{3}{2}$$ (and if set equal to 1, then $$\cos\theta=-\frac{2}{5}$$).

3. $$\displaystyle -\cot\theta-\cos^2\theta$$.
1. $$ \frac{\sqrt{2}}{2} $$ 2. $$ -\frac{1}{\cos\theta} - \frac{3}{2} $$ (If set equal to 1, then $$ \cos\theta = -\frac{2}{5} $$) 3. $$ -\cot\theta - \cos^2\theta $$

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