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Question

Prove that : 1sin103cos10=4

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Solution

LHS :
1sin10o3cos10o

=2[12sin10o32cos10o]

=2[sin30osin10ocos30ocos10o]

=2[sin30ocos10ocos30osin10osin10ocos10o]

=2[sin20osin10ocos10o]

=2×2[sin10ocos10osin10ocos10o]

=4 = RHS

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