Question - Trigonometric Identity Proof Involving Sine and Cosine

Step 1: Write \(1 - \sin^2\theta\) as \(\cos^2\theta\).

\(\cos^2\theta = 1 - \sin^2\theta\)

Step 2: Use the Pythagorean identity, \(1 + \tan^2\theta = \sec^2\theta\), to express \(\cos^2\theta\) as \(1 / \sec^2\theta\).

\(\cos^2\theta = 1 / \sec^2\theta\)

Step 3: Square \(\csc\theta - \sin\theta = a^3\) to get \(\csc^2\theta - 2\csc\theta\sin\theta + \sin^2\theta\).

\((\csc\theta - \sin\theta)^2 = (\csc^2\theta - 2\csc\theta\sin\theta + \sin^2\theta) = a^6\)

Step 4: Use the identity \(\csc\theta\sin\theta = 1\) to simplify.

\(a^6 = \csc^2\theta - 2 + \sin^2\theta\)

Step 5: Express \(\csc^2\theta\) and \(\sin^2\theta\) in terms of \(a^3\).

\(a^6 = (1 + \sin^2\theta) - 2 + \sin^2\theta = 2\sin^2\theta - 1\), since \(\csc^2\theta = 1 + \sin^2\theta\).

Step 6: Express \(\sin^2\theta\) as \(1 - \cos^2\theta\).

\(a^6 = 2(1 - \cos^2\theta) - 1 = 2 - 2\cos^2\theta - 1 = 1 - 2\cos^2\theta\)

Step 7: Repeat similar steps for \(\sec\theta - \cos\theta = b^3\), to find \(b^6\).

\(b^6 = 2\cos^2\theta - 1\)

Step 8: Adding \(a^6\) and \(b^6\).

\(a^6 + b^6 = (1 - 2\cos^2\theta) + (2\cos^2\theta - 1) = 0\)

Step 9: Find \(3a^2b^2(a^2 + b^2)\).

Recognize that \(a^3 = \csc\theta - \sin\theta\) and \(b^3 = \sec\theta - \cos\theta\), and square both.

\(a^2 = (\csc\theta - \sin\theta)^{2/3}\) and \(b^2 = (\sec\theta - \cos\theta)^{2/3}\)

Calculate \(a^2 + b^2\).

\(a^2 + b^2 = (\csc\theta - \sin\theta)^{2/3} + (\sec\theta - \cos\theta)^{2/3}\)

\(a^2 + b^2 = (1 + \sin^2\theta)^{2/3} + (1 + \cos^2\theta)^{2/3}\) (using \(\csc^2\theta = 1 + \sin^2\theta\) and \(\sec^2\theta = 1 + \cos^2\theta\))

\(a^2 + b^2 = (1/\sin^2\theta + \sin^2\theta) + (1/\cos^2\theta + \cos^2\theta)\) (simplifying)

Now use \(1/\sin^2\theta = \csc^2\theta\) and \(1/\cos^2\theta = \sec^2\theta\).

\(a^2 + b^2 = \csc^2\theta + \sin^2\theta + \sec^2\theta + \cos^2\theta\)

\(a^2 + b^2 = 1 + \sin^2\theta + \sin^2\theta + 1 + \cos^2\theta + \cos^2\theta\)

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

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

\(a^2 + b^2 = 2 + 2(1) = 4\)

Now, calculate \(3a^2b^2(a^2 + b^2)\),

\(3a^2b^2(a^2 + b^2) = 3(1)\)

\(3a^2b^2(a^2 + b^2) = 3\)

Finally, the sum is:

\(a^6 + b^6 + 3a^2b^2(a^2 + b^2) = 0 + 3 = 1\)