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

Prove that (sin θ + cosec θ)/(sin θ) = 2 + cot^2θ.

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प्रश्न

Prove that `(sin θ + "cosec"  θ)/(sin θ) = 2 + cot^2θ`.

सिद्धांत
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उत्तर

L.H.S. = `(sin θ + "cosec"  θ)/(sin θ)`

= `(sin θ)/(sin θ) + ("cosec"  θ)/(sin θ)`

= 1 + cosec θ × cosec θ   ...`[∵ "cosec"  θ = 1/(sin θ)]`

= 1 + cosec2θ

= 1 + 1 + cot2θ   ...[∵ 1 + cot2θ = cosec2θ]

= 2 + cot2θ

= R.H.S.

∴ `(sin θ + "cosec"  θ)/(sin θ) = 2 + cot^2θ`

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पाठ 6: Trigonometry - Exercise

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 Evaluate sin25° cos65° + cos25° sin65°


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Activity:

L.H.S. = `square`

= `square (1 - (sin^2θ)/(tan^2θ))`

= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`

= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`

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= R.H.S.


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