Let $x t a n α + y s i n α = α, α x c o s e c α + y c o s α = 1$ be two straight lines. Let P be the point of intersection of the lines in the limiting position when \alpha \rightarrow 0, if the point P is (h, k), then |h + k| is___.

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Solution

$P (\frac{α c o s α - s i n α}{s i n α - α}, \frac{s i n^{2} α - α^{2} c o s α}{(s i n α - α) s i n α c o s α})$
As limit $α \to 0^{+}$
$P \equiv (2, - 1)$
$∴ h + k = 1$

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Let xtanα+ysinα=α,αxcosecα+ycosα=1 be two straight lines. Let P be the point of intersection of the lines in the limiting position when \alpha \rightarrow 0, if the point P is (h, k), then |h + k| is___.

P(αcosα−sinαsinα−α,sin2α−α2cosα(sinα−α)sinαcosα) As limit α→0+ P≡(2,−1) ∴h+k=1

Let $x t a n α + y s i n α = α, α x c o s e c α + y c o s α = 1$ be two straight lines. Let P be the point of intersection of the lines in the limiting position when \alpha \rightarrow 0, if the point P is (h, k), then |h + k| is___.

$P (\frac{α c o s α - s i n α}{s i n α - α}, \frac{s i n^{2} α - α^{2} c o s α}{(s i n α - α) s i n α c o s α})$
As limit $α \to 0^{+}$
$P \equiv (2, - 1)$
$∴ h + k = 1$