tan²x+secx=5 , then sec x=
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Given that sec x + tan x = 5, or
1+sin x = 5 cos x
Consider a right angled triangle ABC, with AB as the adjacent and BC as the opposite and AC as the hypotenuse and <A = x.
So sec x + tan x = 5, or
sec A + tan A = 5, or
AC/AB + BC/AB = 5, or
AC+BC = 5AB, or
AC = 5AB-BC
AB^2+BC^2=AC^2 = [5AB-BC]^2 = 25AB^2–10AB.BC+BC^2, or
AB^2 = 25AB^2–10AB.BC, or
AB = 25AB-10BC, or
24AB = 10BC, or
12AB = 5BC, or
BC/AB = 12/5 or tan A = 12/5
Hence sec A = 5–12/5= 13/5.
sec x = 13/5.
The RAT has a base of 5 units, opposite of 12 units and a hypotenuse of 13 units.
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