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Prove the Following.Sec2θ + Cosec2θ = Sec2θ × Cosec2θ - Geometry

Question

Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ 

Solution

\[\sec^2 \theta + {cosec}^2 \theta\]
\[ = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}\]
\[ = \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta \sin^2 \theta}\]
\[ = \frac{1}{\cos^2 \theta \sin^2 \theta}\]
\[ = \frac{1}{\cos^2 \theta} \times \frac{1}{\sin^2 \theta}\]
\[ = \sec^2 \theta  \text{ cosec }^2 \theta\]
  Is there an error in this question or solution?
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Prove the Following.Sec2θ + Cosec2θ = Sec2θ × Cosec2θ Concept: Application of Trigonometry.
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