Let’s expand the expression:

\( = 1(1 + \cot \theta – \csc \theta) + \tan \theta(1 + \cot \theta – \csc \theta) + \sec \theta(1 + \cot \theta – \csc \theta) \)

\( = 1 + \cot \theta – \csc \theta + \tan \theta + \tan \theta \cot \theta – \tan \theta \csc \theta + \sec \theta + \sec \theta \cot \theta – \sec \theta \csc \theta \)

Simplify using trigonometric identities:

\( \tan \theta \cot \theta = 1 \)

\( \tan \theta \csc \theta = \frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\sin \theta} = \frac{1}{\cos \theta} = \sec \theta \)

\( \sec \theta \cot \theta = \frac{1}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta} = \frac{1}{\sin \theta} = \csc \theta \)

Substituting back:

\( = 1 + \cot \theta – \csc \theta + \tan \theta + 1 – \sec \theta + \sec \theta + \csc \theta – \sec \theta \csc \theta \)

Many terms cancel out:

\( = 2 + \cot \theta + \tan \theta – \sec \theta \csc \theta \)

Expand both squares:

\( = (\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta) + (\sin^2 \theta – 2\sin \theta \cos \theta + \cos^2 \theta) \)

Combine like terms:

\( = 2\sin^2 \theta + 2\cos^2 \theta \)

Factor out 2:

\( = 2(\sin^2 \theta + \cos^2 \theta) \)

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):

\( = 2(1) = 2 \)

We know that:

\( \sec^2 \theta – 1 = \tan^2 \theta \)

\( \csc^2 \theta – 1 = \cot^2 \theta \)

So the expression becomes:

\( \tan^2 \theta \cdot \cot^2 \theta \)

Since \( \cot \theta = \frac{1}{\tan \theta} \):

\( = \tan^2 \theta \cdot \frac{1}{\tan^2 \theta} = 1 \)

Start with the left side:

\( (\csc \theta – \cot \theta)^2 \)

Express in terms of sine and cosine:

\( = \left( \frac{1}{\sin \theta} – \frac{\cos \theta}{\sin \theta} \right)^2 = \left( \frac{1 – \cos \theta}{\sin \theta} \right)^2 = \frac{(1 – \cos \theta)^2}{\sin^2 \theta} \)

Using \( \sin^2 \theta = 1 – \cos^2 \theta \):

\( = \frac{(1 – \cos \theta)^2}{(1 – \cos \theta)(1 + \cos \theta)} \)

Cancel \( (1 – \cos \theta) \):

\( = \frac{1 – \cos \theta}{1 + \cos \theta} \)

Which matches the right side.

Start with the left side:

Rationalize the numerator by multiplying numerator and denominator by \( 1 + \sin A \):

\( \sqrt{\frac{(1 + \sin A)^2}{1 – \sin^2 A}} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} = \frac{1 + \sin A}{\cos A} \)

Split the fraction:

\( = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A \)

Which matches the right side.

Start with the left side:

Express everything in terms of tan A:

\( \cot A = \frac{1}{\tan A} \), so \( \cot^2 A = \frac{1}{\tan^2 A} \)

Substitute:

\( \frac{1 – \tan^2 A}{\frac{1}{\tan^2 A} – 1} = \frac{1 – \tan^2 A}{\frac{1 – \tan^2 A}{\tan^2 A}} \)

Divide by a fraction is same as multiplying by its reciprocal:

\( = (1 – \tan^2 A) \cdot \frac{\tan^2 A}{1 – \tan^2 A} = \tan^2 A \)

Which matches the right side.

Start with the left side:

\( \frac{1}{\cos \theta} – \cos \theta = \frac{1 – \cos^2 \theta}{\cos \theta} = \frac{\sin^2 \theta}{\cos \theta} \)

Now, the right side:

\( \tan \theta \cdot \sin \theta = \frac{\sin \theta}{\cos \theta} \cdot \sin \theta = \frac{\sin^2 \theta}{\cos \theta} \)

Both sides are equal.

First, expand the expression:

\( = \sec A (1 – \sin A) \sec A + \sec A (1 – \sin A) \tan A \)

\( = \sec^2 A (1 – \sin A) + \sec A \tan A (1 – \sin A) \)

Factor out \( (1 – \sin A) \):

\( = (1 – \sin A)(\sec^2 A + \sec A \tan A) \)

Express in terms of cosine and sine:

\( = (1 – \sin A)\left( \frac{1}{\cos^2 A} + \frac{\sin A}{\cos^2 A} \right) = (1 – \sin A)\left( \frac{1 + \sin A}{\cos^2 A} \right) \)

Numerator becomes \( 1 – \sin^2 A = \cos^2 A \):

\( = \frac{\cos^2 A}{\cos^2 A} = 1 \)

Expand both squares:

\( = \sin^2 A + 2\sin A \csc A + \csc^2 A + \cos^2 A + 2\cos A \sec A + \sec^2 A \)

Simplify using identities:

\( \sin A \csc A = 1 \), \( \cos A \sec A = 1 \)

\( \csc^2 A = 1 + \cot^2 A \), \( \sec^2 A = 1 + \tan^2 A \)

Also, \( \sin^2 A + \cos^2 A = 1 \)

Substitute:

\( = 1 + 2(1) + (1 + \cot^2 A) + 2(1) + (1 + \tan^2 A) \)

Combine like terms:

\( = 1 + 2 + 1 + \cot^2 A + 2 + 1 + \tan^2 A \)

\( = 7 + \tan^2 A + \cot^2 A \)

First two terms are difference of squares:

\( = (1 – \cos^2 \theta)(1 + \cot^2 \theta) \)

\( = \sin^2 \theta \cdot \csc^2 \theta \) (since \( 1 + \cot^2 \theta = \csc^2 \theta \))

\( = \sin^2 \theta \cdot \frac{1}{\sin^2 \theta} = 1 \)

We know the identity:

\( \sec^2 \theta – \tan^2 \theta = 1 \)

This can be written as:

\( (\sec \theta + \tan \theta)(\sec \theta – \tan \theta) = 1 \)

Given \( \sec \theta + \tan \theta = p \), then:

\( p (\sec \theta – \tan \theta) = 1 \)

Therefore:

\( \sec \theta – \tan \theta = \frac{1}{p} \)

We know the identity:

\( \csc^2 \theta – \cot^2 \theta = 1 \)

This can be written as:

\( (\csc \theta + \cot \theta)(\csc \theta – \cot \theta) = 1 \)

Given \( \csc \theta + \cot \theta = k \), then \( \csc \theta – \cot \theta = \frac{1}{k} \)

Add the two equations:

\( 2\csc \theta = k + \frac{1}{k} = \frac{k^2 + 1}{k} \)

Subtract the two equations:

\( 2\cot \theta = k – \frac{1}{k} = \frac{k^2 – 1}{k} \)

Now, \( \cos \theta = \frac{\cot \theta}{\csc \theta} = \frac{\frac{k^2 – 1}{2k}}{\frac{k^2 + 1}{2k}} = \frac{k^2 – 1}{k^2 + 1} \)

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.