In ∆ABC, ∠C = 90° and AB = c, BC = a, CA = b;

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In ∆ABC, ∠C = 90° and AB = c, BC = a, CA = b; then the value of (cosec B – cos A) is

In ∆ABC, AB² = AC² + BC²
⇒ c² = a² + b² ........ (i)
From ∆ABC,

cosec B =		AB	=	c	....... (ii)
		AC		b

cos A =		AC	=	b
		AB		c

∴ cosec B – cos A =		c	=	b
		b		c

=		c² - b²	=	a²
		bc		bc