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Make K the subject of the relation T sqrt frac ... - WAEC Mathematics 2015 Question

Make K the subject of the relation T = (\sqrt{\frac{TK - H}{K - H}})

A
K = \(\frac{H(T^2 - 1)}{T^2 - T}\)
B
K = \(\frac{HT}{(T - 1)^2}\)
C
K = \(\frac{H(T^2 + 1)}{T}\)
D
K = \(\frac{H(T - 1)}{T}\)
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Correct Option: A

T = (\sqrt{\frac{TK - H}{K - H}})

Taking the square of both sides, give

T2 = (\frac{TK - H}{K - H})

T2(K - H) = TK - H

T2K - T2H = TK - H

T2K - TK = T2H - H

K(T2 - T) = H(T2 - 1)

K = (\frac{H(T^2 - 1)}{T^2 - T})

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