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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Prove the following: sin2A cos2B + cos2 A sin2B + cos2A cos2B + sin2A sin2B = 1

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प्रश्न

Prove the following:  

sin2A cos2B + cos2A sin2B + cos2A cos2B + sin2A sin2B = 1

बेरीज
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उत्तर

L.H.S. = sin2A cos2B + cos2A sin2B + cos2A cos2B + sin2A sin2B

= (sin2A cos2B + sin2A sin2B) + (cos2A sin2B + cos2A cos2B)

= sin2A (cos2B + sin2B) + cos2A (sin2B + cos2B)

= sin2A + cos2A   ...[∵ sin2A + cos2A = 1]

=1

=R.H.S.

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पाठ 2: Trigonometry - 1 - MISCELLANEOUS EXERCISE - 2 [पृष्ठ ३३]

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बालभारती Mathematics and Statistics (Arts and Science) Part 1 [English] Standard 11 Maharashtra State Board
पाठ 2 Trigonometry - 1
MISCELLANEOUS EXERCISE - 2 | Q 10) i) | पृष्ठ ३३

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