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Perform the multiplication a13(cot⁡x+csc⁡x)(cot⁡x−csc⁡x)a^{13} (\cot x + \csc x)(\cot x - \csc x)

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Full solution

Q. Perform the multiplication a13(cot⁡x+csc⁡x)(cot⁡x−csc⁡x)a^{13} (\cot x + \csc x)(\cot x - \csc x)
  1. Identify Pattern: Identify the pattern in the expression.\newlineThe given expression (cot⁡x+csc⁡x)(cot⁡x−csc⁡x)(\cot x + \csc x)(\cot x - \csc x) resembles the difference of squares pattern a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b).
  2. Apply Formula: Apply the difference of squares formula.\newlineUsing the pattern a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b), we can rewrite the expression as cot⁡2x−csc⁡2x\cot^2 x - \csc^2 x.
  3. Simplify Using Identities: Simplify the expression using trigonometric identities.\newlineWe know that cot⁡2x=cos⁡2xsin⁡2x\cot^2 x = \frac{\cos^2 x}{\sin^2 x} and csc⁡2x=1sin⁡2x\csc^2 x = \frac{1}{\sin^2 x}. Therefore, we can write cot⁡2x−csc⁡2x\cot^2 x - \csc^2 x as cos⁡2xsin⁡2x−1sin⁡2x\frac{\cos^2 x}{\sin^2 x} - \frac{1}{\sin^2 x}.
  4. Combine Over Common Denominator: Combine the terms over a common denominator.\newlineSince both terms have the same denominator sin⁡2x\sin^2 x, we can combine them to get (cos⁡2x)−1sin⁡2x\frac{(\cos^2 x) - 1}{\sin^2 x}.
  5. Use Pythagorean Identity: Use the Pythagorean identity to simplify the numerator.\newlineThe Pythagorean identity states that cos⁡2x+sin⁡2x=1\cos^2 x + \sin^2 x = 1. Therefore, cos⁡2x−1=−(sin⁡2x)\cos^2 x - 1 = -(\sin^2 x). We can substitute this into our expression to get −((sin⁡2x)/(sin⁡2x))-((\sin^2 x)/(\sin^2 x)).
  6. Final Simplification: Simplify the expression.\newlineSince sin⁡2x\sin^2 x in the numerator and denominator are the same, they cancel out, leaving us with −1-1.

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