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question no. 10
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SinA= a/√(4b^2-3a^2)
CosA= 2√(b^2-a^2)/√(4b^2-3a^2)
TanA= a/2√(b^2-a^2)
CosA= 2√(b^2-a^2)/√(4b^2-3a^2)
TanA= a/2√(b^2-a^2)
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It will help you.........
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ChelsiNegi:
tnx
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