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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