InterviewSolution
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InterviewSolution
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solve the following equation `sec^(-1).(x)/(a) - sec^(-1).(x)/(b) = sec^(-1) b - sec^(-1) a, a ge 1, b ge 1, a!= b` |
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Answer» Correct Answer - `x = ab` `sec^(-1).(x)/(a) - sec^(-1).(x)/(b) = sec^(-1) b - sec^(-1) a` `rArr cos^(-1).(a)/(x) - cos^(-1).(b)/(x) = cos^(-1).(1)/(b) - cos^(-1).(1)/(a)` `rArr cos^(-1).(a)/(x) + cos^(-1).(1)/(a) = cos^(-1).(b)/(x) - cos^(-1).(1)/(b)` `rArr cos^(-1) [(1)/(x) - sqrt(1 - (a^(2))/(x^(2))) sqrt(1 - (1)/(a^(2)))] = cos^(-1) [(1)/(x) - sqrt(1 - (b^(2))/(x^(2))) sqrt(1 -(1)/(b^(2)))]` `rArr (1)/(x) - sqrt(1 - (1)/(a^(2)) - (a^(2))/(x^(2)) + (1)/(x^(2))) = (1)/(x) - sqrt(1 - (b^(2))/(x^(2)) - (1)/(b^(2)) + (1)/(x^(2)))` `rArr sqrt(1 -(1)/(a^(2)) - (a^(2))/(x^(2)) + (1)/(x^(2))) = sqrt(1 - (1)/(b^(2)) - (b^(2))/(x^(2)) + (1)/(x^(2)))` `rArr -(1)/(a^(2)) - (a^(2))/(x^(2)) = -(1)/(b^(2)) - (b^(2))/(x^(2))` `rArr (1)/(b^(2)) - (1)/(a^(2)) = (a^(2) - b^(2))/(x^(2))` `rArr x^(2) = a^(2) b^(2)` `:. x = +- ab` But `x = 0 ab` does not satisfy the given equation Hence, `x = ab` is the required solution |
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