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Cosec square theta+ root 3 cot theta -7=0 |
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Answer» we KNOW, 1 + cot^2 theta + √3 cot theta - 7=0 let cot theta = x (x)^2 + √3x - 6 = 0 ↪(x )^2 +2√3x - √3x - 6 = 0 ↪x ( x + 2√3 ) - √3 ( x + 2√3 ) = 0 ↪( x - √3 ) ( x + 2√3 ) ↪ x = √3 OR x = -2√3 Therefor cot theta = √3 OR cot theta = -2/√3 |
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