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About dhol in assamese |
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Answer» °-θ) = COS θcos (90°-θ) = sin θtan (90°-θ) = cot θcsc (90°-θ) = SEC θsec (90°-θ) = CSC θcot (90°-θ) = TAN θsin (90°+θ) = cos θcos (90°+θ) = -sin θtan (90°+θ) = -cot θcsc (90°+θ) = sec θsec (90°+θ) = -csc θcot (90°+θ) = -tan θsin (180°-θ) = sin θcos (180°-θ) = -cos θtan (180°-θ) = -tan θcsc (180°-θ) = csc θsec (180°-θ) = -sec θcot (180°-θ) = -cot θsin (180°+θ) = -sin θcos (180°+θ) = -cos θtan (180°+θ) = tan θcsc (180°+θ) = -csc θsec (180°+θ) = -sec θcot (180°+θ) = cot θsin (270°-θ) = -cos θcos (270°-θ) = -sin θtan (270°-θ) = cot θcsc (270°-θ) = -sec θsec (270°-θ) = -csc θcot (270°-θ) = tan θsin (270°+θ) = -cos θcos (270°+θ) = sin θtan (270°+θ) = -cot θcsc (270°+θ) = -sec θsec (270°+θ) = cos θcot (270°+θ) = -tan θ |
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