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Sin8AcosA-sin6Acos3A/cos2AcosA-sin3Asin4A=sin2A/cos2A

Sin8AcosA-sin6Acos3A/cos2AcosA-sin3Asin4A=sin2A/cos2A

Grade:11

1 Answers

Anish Singhal
askIITians Faculty 1194 Points
2 years ago
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check this solution, i think you should check question once again

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