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Direction Cosines

If x = 4aba...

Question

If $x = \frac{4 a b}{a + b}$ , then prove that $\frac{x + 2 a}{x - 2 a} + \frac{x + 2 b}{x - 2 b} = 2$

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Solution

Given: $x = \frac{4 a b}{a + b}$
$\Rightarrow \frac{x}{2 a} = \frac{2 b}{a + b}$
by componendo and dividendo
$\Rightarrow \frac{x + 2 a}{x - 2 a} = \frac{2 b + a + b}{2 b - (a + b)}$
$\Rightarrow \frac{x + 2 a}{x - 2 a} = \frac{3 b + a}{(b - a)} . . . . (i)$
again $x = \frac{4 a b}{a + b}$
$\Rightarrow \frac{x}{2 b} = \frac{2 b}{a + b}$
by componendo and dividendo
$\frac{x + 2 b}{x - 2 b} = \frac{2 a + a + b}{2 a - (a + b)}$
$\Rightarrow \frac{x + 2 b}{x - 2 b} = \frac{3 a + b}{(a - b)}$
$\Rightarrow \frac{x + 2 b}{x - 2 b} = \frac{- (3 a + b)}{b - a} . . . . (i i)$
On adding (i) and (ii)
$\frac{x + 2 a}{x - 2 a} + \frac{x + 2 b}{x - 2 b} = \frac{3 b + a}{b - a} - \frac{(3 a + b)}{(b - a)}$
$\Rightarrow \frac{x + 2 a}{x - 2 a} + \frac{x + 2 b}{x - 2 b} = \frac{3 b + a - 3 a - b}{(b - a)}$
$\Rightarrow \frac{x + 2 a}{x - 2 a} + \frac{x + 2 b}{x - 2 b} = \frac{2 b - 2 a}{b - a}$
$\Rightarrow \frac{x + 2 a}{x - 2 a} + \frac{x + 2 b}{x - 2 b} = \frac{2 (b - a)}{(b - a)}$
$\Rightarrow \frac{x + 2 a}{x - 2 a} + \frac{x + 2 b}{x - 2 b} = 2$

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Similar questions

x = \frac{4 a b}{a + b}

, let us show that

\frac{x + 2 a}{x - 2 a} + \frac{x + 2 b}{x - 2 b} = 2

a \neq 0, b \neq 0

a \neq b

x = \frac{4 a b}{a + b}

\frac{x + 2 a}{x - 2 a} + \frac{x + 2 b}{x - 2 b}

x = \frac{4 a b}{a + b}

a \neq b \neq 0

, then the value of

\frac{x + 2 a}{x - 2 a} + \frac{x + 2 b}{x - 2 b}

is
(Assume that everything is defined)

x, y :

(a + 2 b) x + (2 a - b) y = 2, (a - 2 b) x + (2 a + b) y = 3

.

x

y

:

(a + 2 b) x + (2 a - b) y = 2, (a - 2 b) x + (2 a + b) y = 3

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