given sina = -sprt7/4, calculate tana+cota

Given that \( \sin(a) = -\frac{\sqrt{7}}{4} \), we can calculate \( \tan(a) + \cot(a) \).

### Step 1: Calculate \( \cos(a) \)
Using the Pythagorean identity:

\[
\sin^2(a) + \cos^2(a) = 1
\]

Substitute \( \sin(a) = -\frac{\sqrt{7}}{4} \):

\[
\left(-\frac{\sqrt{7}}{4}\right)^2 + \cos^2(a) = 1
\]

\[
\frac{7}{16} + \cos^2(a) = 1
\]

\[
\cos^2(a) = 1 - \frac{7}{16} = \frac{16}{16} - \frac{7}{16} = \frac{9}{16}
\]

\[
\cos(a) = \pm \frac{3}{4}
\]

Since \( \sin(a) \) is negative, the quadrant must be III or IV, where \( \cos(a) \) is also negative:

\[
\cos(a) = -\frac{3}{4}
\]

### Step 2: Calculate \( \tan(a) \) and \( \cot(a) \)
\[
\tan(a) = \frac{\sin(a)}{\cos(a)} = \frac{-\frac{\sqrt{7}}{4}}{-\frac{3}{4}} = \frac{\sqrt{7}}{3}
\]

\[
\cot(a) = \frac{1}{\tan(a)} = \frac{3}{\sqrt{7}}
\]

### Step 3: Calculate \( \tan(a) + \cot(a) \)
\[
\tan(a) + \cot(a) = \frac{\sqrt{7}}{3} + \frac{3}{\sqrt{7}}
\]

To simplify, find a common denominator:

\[
\tan(a) + \cot(a) = \frac{\sqrt{7} \times \sqrt{7} + 3 \times 3}{3\sqrt{7}} = \frac{7 + 9}{3\sqrt{7}} = \frac{16}{3\sqrt{7}}
\]

Rationalize the denominator:

\[
\tan(a) + \cot(a) = \frac{16\sqrt{7}}{21}
\]

Thus, the final expression is:

\[
\tan(a) + \cot(a) = \frac{16\sqrt{7}}{21}
\]

Would you like more details or have any questions?

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Here are 5 related questions you might find interesting:
1. What is the value of \( \tan(a) + \cot(a) \) when \( \sin(a) = \frac{3}{5} \)?
2. How does the sign of \( \cos(a) \) affect the value of \( \tan(a) + \cot(a) \)?
3. Can you solve \( \tan(a) + \cot(a) \) directly using trigonometric identities?
4. How would the value change if \( \sin(a) \) was positive instead of negative?
5. What other trigonometric identities involve both \( \tan(a) \) and \( \cot(a) \)?

**Tip:** When working with trigonometric functions, always consider the sign and the quadrant to correctly determine the values.

Learn how to calculate tan(a) + cot(a) for sin(a) = -sqrt(7)/4 using trigonometric identities and the Pythagorean identity. Understand the step-by-step process and rationale behind the solution. Suitable for high school students studying trigonometry.

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