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महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

5/(sin^2θ) – 5cot^2θ, complete the activity given below. Activity: 5/(sin^2θ) – 5cot^2θ = square (1/(sin^2θ) – cot^2θ) = 5(square – cot^2θ) ...[1/(sin^2θ) = square] = 5(1) = square

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प्रश्न

`5/(sin^2θ) - 5cot^2θ`, complete the activity given below.

Activity:

`5/(sin^2θ) - 5cot^2θ`

= `square (1/(sin^2θ) - cot^2θ)`

= `5(square - cot^2θ)   ...[1/(sin^2θ) = square]`

= 5(1)

= `square`

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उत्तर

`5/(sin^2θ) - 5cot^2θ`

= \[\boxed{5}\left(\frac{1}{\text{sin}^2θ} - \text{cot}^2θ\right)\]

= \[5\left(\boxed{\text{cosec}^2θ} - \text{cot}^2θ\right)\]   \[...[\frac{1}{\text{sin}^2θ} = \boxed{\text{cosec}^2θ}]\]

= 5(1)

= \[\boxed{5}\]

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पाठ 6: Trigonometry - Exercise

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