if m , n are the solutions of: a tanx+b secx=c,then tan(m+n)=?

Question

1) 2ac/a^2-c^2
2)2ac/c^2-a^2
plse xplain ur ans

Answer 1

Like DutchProf, I find only one of the two answers Nishi says she is looking for. I present only an outline below -- you need to fill in some algebra steps for yourself, as well as spacing the work out a bit better and leaving out my rambling commentary!

Rearrange the equation as b sec x = c - a tan x. Then square both sides and replace (sec x)^2 by 1 + (tan x)^2, then rearrange the terms to present this quadratic equation as
(a^2 - b^2)(tan x)^2 + 2ac tan x + c^2 - b^2 = 0. The sum of the roots (i.e. tan m + tan n) is -2ac/(a^2 - b^2), while the product is (c^2 - b^2)/(a^2 - b^2).
Now tan (m + n) = (tan m + tan n)/(1 - (tan m)(tan n))

which is (-2ac/(a^2 - b^2))/(1 - (c^2 - b^2)/(a^2 -b^2)).

Multiply numerator and denominator by a^2 - b^2, giving
-2ac/(a^2 - b^2 - (c^2 - b^2)), which reduces to

-2ac/(a^2 - c^2); i.e. 2ac/(c^2 - a^2)

Answer 2

a tan x + b sec x = c
a (sin x / cos x) + b / cos x = c

multiply with cos x

a sin x = b - c cos x
a sin x + c cos x = b

Let h = sqrt (a^2 + c^2) and t = inv tan (c/a). Then

h sin (x + t) = b
x = inv sin (b/h) - t or x = pi/2 - inv sin (b/h) - t

These are the solutions m and n. Their sum

m + n = -2 t

tan (m + n) = tan (-2t) = -2 tan t / (1 - tan t^2)

= - 2 (c/a) / (1 - [c/a]^2)
= - 2ca / (a^2 - c^2)
= 2ac / (c^2 - a^2)

Answer 3

Dutch Prof,

Your insight is good, but you made a couple a careless algebra mistakes. Shame on you.