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Prove the Following.Sec2θ + Cosec2θ = Sec2θ × Cosec2θ - Geometry Mathematics 2

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प्रश्न

Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ 

योग
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उत्तर

\[\sec^2 \theta + {cosec}^2 \theta\]
\[ = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}\]
\[ = \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta \sin^2 \theta}\]
\[ = \frac{1}{\cos^2 \theta \sin^2 \theta}\]
\[ = \frac{1}{\cos^2 \theta} \times \frac{1}{\sin^2 \theta}\]
\[ = \sec^2 \theta  \text{ cosec }^2 \theta\]
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अध्याय 6: Trigonometry - Problem Set 6 [पृष्ठ १३८]

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बालभारती Mathematics 2 [English] Standard 10 Maharashtra State Board
अध्याय 6 Trigonometry
Problem Set 6 | Q 5.03 | पृष्ठ १३८
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