$$ \frac { \ln ( 90 ^ { \circ } - heta ) \cdot \cos 480 ^ { \circ } + \cos ( 180 ^ { \circ } - heta ) \cdot an 4 \sin 390 ^ { \circ } - an 180 ^ { \circ } } { \cos } $$

Question

$$ \frac { \ln ( 90 ^ { \circ } - 	heta ) \cdot \cos 480 ^ { \circ } + \cos ( 180 ^ { \circ } - 	heta ) \cdot 	an 4 \sin 390 ^ { \circ } - 	an 180 ^ { \circ } } { \cos } $$

UpStudy AI · Accepted Answer

We wish to simplify the expression
$$
\frac{\ln\bigl(90^\circ-\theta\bigr)\cdot \cos480^\circ+\cos\bigl(180^\circ-\theta\bigr)\cdot \tan\Bigl(4\sin390^\circ\Bigr)-\tan180^\circ}{\cos\theta}\,,
$$
assuming that the denominator is $$\cos\theta$$ (that is, the original “$$\cos$$” has argument $$\theta$$) and that the factor “$$\tan4 \sin390^\circ$$” is to be read as $$\tan\Bigl(4\sin390^\circ\Bigr)$$.

We now simplify step by step.

---

**1. Evaluate the standard trigonometric values.**

(a) Compute $$\cos480^\circ$$:
$$
480^\circ-360^\circ=120^\circ\quad\text{and}\quad\cos120^\circ=-\frac{1}{2}\,.
$$
Thus,
$$
\cos480^\circ=-\frac{1}{2}\,.
$$

(b) Compute $$\tan180^\circ$$:
$$
\tan180^\circ=0\,.
$$

(c) Compute $$\sin390^\circ$$:
$$
390^\circ-360^\circ=30^\circ\quad\text{and}\quad\sin30^\circ=\frac{1}{2}\,.
$$
Thus,
$$
\sin390^\circ=\frac{1}{2}\,.
$$

Then,
$$
4\sin390^\circ=4\cdot\frac{1}{2}=2\,,
$$
so that
$$
\tan\Bigl(4\sin390^\circ\Bigr)=\tan2\,.
$$
*Note:* The angle “2” here is understood as in the same units that the inner functions imply (when mixing degrees and real numbers one must be cautious; here we leave the result as $$\tan2$$).

(d) Express $$\cos(180^\circ-\theta)$$ in a simpler form. Using the identity
$$
\cos(180^\circ-x)=-\cos x\,,
$$
with $$x=\theta$$ we obtain:
$$
\cos(180^\circ-\theta)=-\cos\theta\,.
$$

---

**2. Substitute the computed values into the numerator.**

The numerator is
$$
\ln\bigl(90^\circ-\theta\bigr)\cdot \cos480^\circ+\cos\bigl(180^\circ-\theta\bigr)\cdot \tan\Bigl(4\sin390^\circ\Bigr)-\tan180^\circ\,.
$$
Substitute the computed values:
- Replace $$\cos480^\circ$$ with $$-\frac{1}{2}$$.
- Replace $$\cos(180^\circ-\theta)$$ with $$-\cos\theta$$.
- Replace $$\tan\Bigl(4\sin390^\circ\Bigr)$$ with $$\tan2$$.
- Replace $$\tan180^\circ$$ with $$0$$.

Thus the numerator becomes
$$
\ln(90^\circ-\theta)\cdot\Bigl(-\frac{1}{2}\Bigr)+\Bigl(-\cos\theta\Bigr)\cdot\tan2-0\,.
$$
That is,
$$
-\frac{1}{2}\ln(90^\circ-\theta)-\cos\theta\,\tan2\,.
$$

---

**3. Write the full expression and simplify.**

The full expression is now
$$
\frac{-\frac{1}{2}\ln(90^\circ-\theta)-\cos\theta\,\tan2}{\cos\theta}\,.
$$
We can split the fraction into two parts:
$$
-\frac{1}{2}\frac{\ln(90^\circ-\theta)}{\cos\theta}-\frac{\cos\theta\,\tan2}{\cos\theta}\,.
$$
Cancelling $$\cos\theta$$ in the second term:
$$
-\frac{1}{2}\frac{\ln(90^\circ-\theta)}{\cos\theta}-\tan2\,.
$$

---

**Final answer:**

$$
-\frac{1}{2}\frac{\ln(90^\circ-\theta)}{\cos\theta}-\tan2\,.
$$
$$ -\frac{1}{2}\frac{\ln(90^\circ-\theta)}{\cos\theta}-\tan2 $$

\( \frac { \ln ( 90 ^ { \circ } - \theta ) \cdot \cos 480 ^ { \circ } + \cos ( 180 ^ { \circ } - \theta ) \cdot \tan 4 \sin 390 ^ { \circ } - \tan 180 ^ { \circ } } { \cos } \)

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