We know that \(\cot(90^\circ – \theta) = \tan\theta\), so \(\cot 54^\circ = \tan 36^\circ\).

Therefore, \(\frac{\tan 36^\circ}{\cot 54^\circ} = \frac{\tan 36^\circ}{\tan 36^\circ} = 1\)

We know that \(\sin(90^\circ – \theta) = \cos\theta\), so \(\sin 78^\circ = \cos 12^\circ\).

Therefore, \(\cos 12^\circ – \sin 78^\circ = \cos 12^\circ – \cos 12^\circ = 0\)

We know that \(\sec(90^\circ – \theta) = \csc\theta\), so \(\sec 59^\circ = \csc 31^\circ\).

Therefore, \(\csc 31^\circ – \sec 59^\circ = \csc 31^\circ – \csc 31^\circ = 0\)

We know that \(\sec\theta = \frac{1}{\cos\theta}\) and \(\cos(90^\circ – \theta) = \sin\theta\).

\(\sec 75^\circ = \frac{1}{\cos 75^\circ} = \frac{1}{\sin 15^\circ}\)

Therefore, \(\sin 15^\circ \sec 75^\circ = \sin 15^\circ \times \frac{1}{\sin 15^\circ} = 1\)

We know that \(\tan(90^\circ – \theta) = \cot\theta\), so \(\tan 64^\circ = \cot 26^\circ\).

Also, \(\cot\theta = \frac{1}{\tan\theta}\), so \(\tan 26^\circ \tan 64^\circ = \tan 26^\circ \cot 26^\circ = \tan 26^\circ \times \frac{1}{\tan 26^\circ} = 1\)

We can rearrange the terms:

\(\tan 48^\circ \tan 42^\circ \tan 16^\circ \tan 74^\circ\)

We know that \(\tan(90^\circ – \theta) = \cot\theta\), so:

\(\tan 42^\circ = \cot 48^\circ\) and \(\tan 74^\circ = \cot 16^\circ\)

Now the expression becomes:

\(\tan 48^\circ \cot 48^\circ \tan 16^\circ \cot 16^\circ\)

Since \(\tan\theta \cot\theta = 1\), the expression simplifies to \(1 \times 1 = 1\)

We know that \(\cos(90^\circ – \theta) = \sin\theta\), so \(\cos 54^\circ = \sin 36^\circ\).

Similarly, \(\sin 54^\circ = \cos 36^\circ\).

Substituting these values:

\(\cos 36^\circ \sin 36^\circ – \sin 36^\circ \cos 36^\circ = 0\)

We know that \(\cot(90^\circ – \theta) = \tan\theta\), so we can write:

\(\tan 2A = \cot(A – 18^\circ) = \tan(90^\circ – (A – 18^\circ)) = \tan(108^\circ – A)\)

Since \(2A\) is acute, both sides are equal:

\(2A = 108^\circ – A\)

\(3A = 108^\circ\)

\(A = 36^\circ\)

We know that \(\cot B = \tan(90^\circ – B)\), so:

\(\tan A = \tan(90^\circ – B)\)

Since both \(A\) and \(B\) are acute angles, we can equate the angles:

\(A = 90^\circ – B\)

Therefore, \(A + B = 90^\circ\)

In any triangle, the sum of interior angles is \(180^\circ\):

\(A + B + C = 180^\circ\)

Therefore, \(A + B = 180^\circ – C\)

Divide both sides by 2:

\(\frac{A + B}{2} = 90^\circ – \frac{C}{2}\)

Now take the tangent of both sides:

\(\tan\left(\frac{A + B}{2}\right) = \tan\left(90^\circ – \frac{C}{2}\right) = \cot\frac{C}{2}\)

This completes the proof.

We can express both terms using complementary angle identities:

\(\sin 75^\circ = \cos 15^\circ\) (since \(\sin\theta = \cos(90^\circ – \theta)\))

\(\cos 65^\circ = \sin 25^\circ\) (since \(\cos\theta = \sin(90^\circ – \theta)\))

Therefore, \(\sin 75^\circ + \cos 65^\circ = \cos 15^\circ + \sin 25^\circ\)

Note: Both 15° and 25° are between 0° and 45° as required.