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If tan x + ...

Question

If $tan x + tan y = a$ and $cot y + cot x = b$ , then $\frac{1}{a} - \frac{1}{b} =$

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Solution

Given : $t a n x + t a n y = a c o t x + c o t y = b$

To find: $\frac{1}{a} - \frac{1}{b}$

$\frac{1}{a} - \frac{1}{b} = \frac{b - a}{a b}$

$= \frac{c o t x + c o t y - (t a n x + t a n y)}{(c o t x + c o t y) (t a n x + t a n y)}$

$= \frac{c o t x - t a n x + c o t y - t a n y}{c o t x t a n x + c o t x t a n y + c o t y t a n x + t a n y c o t y}$

$= \frac{1 / t a n x - t a n x + 1 / t a n y - t a n y}{1 + c o t x t a n y + c o t y t a n x + 1}$

$= \frac{(1 - t a n^{2} x) / t a n x + (10 t a n^{2} y) / t a n y}{2 + t a n y / t a n x + t a n x / t a n y}$

$= \frac{(s e c^{2} x) t a n y + s e c^{2} y t a n x}{2 t a n x / t a n y + t a n^{2} y + t a n^{2} x}$

$= \frac{t a n y - t a n^{2} x t a n y + t a n x - t a n^{2} t t a n x}{(t a n x + t a n y)^{2}}$

$= \frac{(t a n x + t a n y) (1 - t a n x t a n y)}{(t a n x + t a n y)^{2}}$

$= \frac{1 - t a n x t a n y}{t a n x + t a n y}$

$= \frac{1}{t a n (x + y)}$

$= c o t (x + y)$

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