#### Question

Prove the following.

sec^{2}θ + cosec^{2}θ = sec^{2}θ × cosec^{2}θ

#### Solution

\[\sec^2 \theta + {cosec}^2 \theta\]

\[ = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}\]

\[ = \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta \sin^2 \theta}\]

\[ = \frac{1}{\cos^2 \theta \sin^2 \theta}\]

\[ = \frac{1}{\cos^2 \theta} \times \frac{1}{\sin^2 \theta}\]

\[ = \sec^2 \theta \text{ cosec }^2 \theta\]

\[ = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}\]

\[ = \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta \sin^2 \theta}\]

\[ = \frac{1}{\cos^2 \theta \sin^2 \theta}\]

\[ = \frac{1}{\cos^2 \theta} \times \frac{1}{\sin^2 \theta}\]

\[ = \sec^2 \theta \text{ cosec }^2 \theta\]

Is there an error in this question or solution?

Solution Prove the Following.Sec2θ + Cosec2θ = Sec2θ × Cosec2θ Concept: Application of Trigonometry.