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The image contains two parts of a problem related to trigonometric identities and expressions:

1. **Given:** \(\sec x = k\), where \(-90^\circ < x < 90^\circ\), express in terms of \(k\):
   - \(\sin(90^\circ - x)\)
   - \(\tan 2x\)

### Let's solve each part step-by-step:

#### (i) \(\sin(90^\circ - x)\)
We can use the co-function identity for sine:
\[
\sin(90^\circ - x) = \cos(x)
\]
Given that \(\sec x = k\), we know:
\[
\sec x = \frac{1}{\cos x}
\]
Thus:
\[
\cos x = \frac{1}{k}
\]
Therefore:
\[
\sin(90^\circ - x) = \cos x = \frac{1}{k}
\]

#### (ii) \(\tan 2x\)
We can use the double-angle identity for tangent:
\[
\tan 2x = \frac{2\tan x}{1 - \tan^2 x}
\]
Since \(\sec x = k\), and \(\sec x = \frac{1}{\cos x}\), we have:
\[
\cos x = \frac{1}{k}, \quad \text{and} \quad \sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \frac{1}{k^2}}
\]
Now, \(\tan x = \frac{\sin x}{\cos x}\), so:
\[
\tan x = \frac{\sqrt{1 - \frac{1}{k^2}}}{\frac{1}{k}} = k\sqrt{1 - \frac{1}{k^2}} = \sqrt{k^2 - 1}
\]
Substitute \(\tan x = \sqrt{k^2 - 1}\) into the double-angle formula:
\[
\tan 2x = \frac{2\sqrt{k^2 - 1}}{1 - (k^2 - 1)} = \frac{2\sqrt{k^2 - 1}}{2 - k^2}
\]

So, the expressions in terms of \(k\) are:
1. \(\sin(90^\circ - x) = \frac{1}{k}\)
2. \(\tan 2x = \frac{2\sqrt{k^2 - 1}}{2 - k^2}\)

Would you like more details or have any questions?

### Here are 5 related questions:
1. How would you express \(\cos(2x)\) in terms of \(k\)?
2. What is the value of \(\sin x\) in terms of \(k\)?
3. How would you express \(\tan(90^\circ - x)\) in terms of \(k\)?
4. Can you express \(\cot x\) in terms of \(k\)?
5. How does the value of \(\sec x\) influence the range of \(x\) in trigonometric identities?

### Tip:
Always remember that co-function identities like \(\sin(90^\circ - x) = \cos(x)\) are very useful in simplifying trigonometric expressions, especially when given constraints on the angles.

Given that sec x = k where -90° < x < 90°, express in terms of k: (i) sin(90° - x); (ii) tan 2x.

Solve trigonometric expressions by expressing sin(90° - x) and tan 2x in terms of sec x. This problem involves trigonometric identities, co-function identities, and the double-angle formula. Suitable for high school students in Grades 10-12.

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