If SecA+cosA= √3 then find the value of tanA + cot A
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The value of SecA + cosA is never less than 2. The given question something is not right. Let us assume it is 3 instead of √3.
SecA + Cos A = 3
=> Cos² A - 3 cos A + 1 = 0
=> cosA = [3 +- √(9-4) ]/2 = (3 - √5 )/2
=> SecA = 2/(3-√5) = (3+√5)/2
=> tan² A = (7+3√5) / 2
=> cot² A = 2/(7+3√5) = (7-3√5)/2
(tan A + cotA)² = 7/2 + 7/2 + 2 = 9
Tan A + cot A = +3 or - 3
SecA + Cos A = 3
=> Cos² A - 3 cos A + 1 = 0
=> cosA = [3 +- √(9-4) ]/2 = (3 - √5 )/2
=> SecA = 2/(3-√5) = (3+√5)/2
=> tan² A = (7+3√5) / 2
=> cot² A = 2/(7+3√5) = (7-3√5)/2
(tan A + cotA)² = 7/2 + 7/2 + 2 = 9
Tan A + cot A = +3 or - 3
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