a. Show that \( \cos^2 k = \sin k \):

Solution:

At the point of intersection \( P \):

\( f(k) = g(k) \)

\( \cos k = \tan k \)

Using the identity \( \tan k = \frac{\sin k}{\cos k} \):

\( \cos k = \frac{\sin k}{\cos k} \)

Multiply both sides by \( \cos k \):

\( \cos^2 k = \sin k \)

Thus, we have shown the required identity.

b. Show tangents intersect at right angles:

Solution:

First, find the derivatives:

\( f'(x) = -\sin x \) ⇒ \( f'(k) = -\sin k \)

\( g'(x) = \sec^2 x \) ⇒ \( g'(k) = \sec^2 k = \frac{1}{\cos^2 k} \)

From part (a), we know \( \cos^2 k = \sin k \), so:

\( g'(k) = \frac{1}{\sin k} \)

Product of gradients:

\( f'(k) \times g'(k) = -\sin k \times \left(\frac{1}{\sin k}\right) = -1 \)

Since the product of gradients is \(-1\), the tangents are perpendicular.

c. Find \( \sin k \) in form \( \frac{a + \sqrt{b}}{c} \):

Solution:

From \( \cos^2 k = \sin k \) and \( \cos^2 k = 1 – \sin^2 k \):

\( 1 – \sin^2 k = \sin k \)

Rearrange:

\( \sin^2 k + \sin k – 1 = 0 \)

Let \( u = \sin k \):

\( u^2 + u – 1 = 0 \)

Using quadratic formula:

\( u = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \)

Since \( 0 < k < \frac{\pi}{2} \), we take the positive solution:

\( \sin k = \frac{-1 + \sqrt{5}}{2} = \frac{\sqrt{5} – 1}{2} \)

This is in the required form with \( a = -1 \), \( b = 5 \), \( c = 2 \).