Question

secA+tanA=u prove that tanA=(u^2-1)/2u

Answer 1

Answer:

sec²A - tan²A =1

or, (sec A+tan A)(sec A- tan A)=1

or, sec A - tan A= 1/u .....(1)

&

secA+tanA=u .......(2)

(2)-(1), we get,

2 tan A =u - (1/u)= (u²-1)/u

or, tan A= (u²-1)/2u

Answer 2

secA + tanA = u

squaring both side

(secA + tanA)^2 = u^2

sec^2A + tan^2A + 2 secA tanA = u^2

subtracting 1 from both side

sec^2A - 1 + tan^2A + 2 secA tanA = u^2 - 1

tan^2A + tan^2A + 2 secA tanA = u^2 - 1

2 tan^2A + 2 secA tanA = u^2 - 1

2 tanA ( tanA + secA) = u^2 - 1

2 tanA * u = u^2 - 1

tanA = (u^2 - 1)/2u

proved