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sin 43° cos 47° + sin 47° cos43° is ...........

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Given, sin43°cos47° + cos43°sin47°⇒ sin43°cos(90 - 43)° + cos43°sin(90 - 43)°⇒ sin43°sin43° + cos43°cos43°⇒ sin243° + cos243° [Since, sin2θ + cos2θ = 1]= 1\xa0\xa0
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Jai mahakal
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Sin90
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