Given $$ f(x)=- an ^{2}(x) $$, write the equation of the line tangent to $$ y=f(x) $$ when $$ x=\frac{4 \pi}{3} $$

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Given $$ f(x)=-	an ^{2}(x) $$, write the equation of the line tangent to $$ y=f(x) $$ when $$ x=\frac{4 \pi}{3} $$

UpStudy AI · Accepted Answer

To find the equation of the tangent line to $$ y = f(x) $$ at $$ x = \frac{4\pi}{3} $$, where $$ f(x) = -\tan^2(x) $$, follow these steps:

### 1. Evaluate $$ f(x) $$ at $$ x = \frac{4\pi}{3} $$:

$$
f\left(\frac{4\pi}{3}\right) = -\tan^2\left(\frac{4\pi}{3}\right)
$$

First, find $$ \tan\left(\frac{4\pi}{3}\right) $$:

$$
\tan\left(\frac{4\pi}{3}\right) = \tan\left(\pi + \frac{\pi}{3}\right) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}
$$

Thus,

$$
f\left(\frac{4\pi}{3}\right) = -\left(\sqrt{3}\right)^2 = -3
$$

So, the point of tangency is $$ \left(\frac{4\pi}{3}, -3\right) $$.

### 2. Find the derivative $$ f'(x) $$:

$$
f(x) = -\tan^2(x) \
f'(x) = -2\tan(x) \cdot \sec^2(x)
$$

### 3. Evaluate $$ f'(x) $$ at $$ x = \frac{4\pi}{3} $$:

First, find $$ \sec\left(\frac{4\pi}{3}\right) $$:

$$
\cos\left(\frac{4\pi}{3}\right) = \cos\left(\pi + \frac{\pi}{3}\right) = -\cos\left(\frac{\pi}{3}\right) = -\frac{1}{2} \
\sec\left(\frac{4\pi}{3}\right) = \frac{1}{\cos\left(\frac{4\pi}{3}\right)} = -2 \
\sec^2\left(\frac{4\pi}{3}\right) = 4
$$

Now,

$$
f'\left(\frac{4\pi}{3}\right) = -2 \cdot \sqrt{3} \cdot 4 = -8\sqrt{3}
$$

### 4. Write the equation of the tangent line:

Using the point-slope form of a line:

$$
y - y_1 = m(x - x_1)
$$

Where $$ m = -8\sqrt{3} $$ and $$ (x_1, y_1) = \left(\frac{4\pi}{3}, -3\right) $$:

$$
y - (-3) = -8\sqrt{3} \left(x - \frac{4\pi}{3}\right) \
y + 3 = -8\sqrt{3} \left(x - \frac{4\pi}{3}\right)
$$

### Final Answer:

After simplifying, an equation for the tangent line is

y + 3 = –8 √3 (x – 4π⁄3).
The equation of the tangent line to $$ y = -\tan^2(x) $$ at $$ x = \frac{4\pi}{3} $$ is: $$ y + 3 = -8\sqrt{3}\left(x - \frac{4\pi}{3}\right) $$

Given , write the equation of the line tangent to when

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