Question 176
In ΔABC, ∠C = 90° and AB = c, BC = a, CA = b; then the value of (cosec B - cos A) is
Solution
In $$\triangle$$$$ABC, AB^2 = AC^2 + BC^2$$
=> $$c^2 = a^2 + b^2 => c^2 - b^2 = a^2$$
$$cosecB = \frac{AB}{BC} = \frac{c}{b}$$
$$cosA = \frac{AC}{AB} = \frac{b}{c}$$
$$\therefore cosecB - cosA = \frac{c}{b} - \frac{b}{c}$$
= $$\frac{c^2-b^2}{bc} = \frac{a^2}{bc}$$
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