SSC CGL 30 Aug 2015 Paper

Question : 126 of 200

Marks: +1, -0

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

Solution:

In ΔABC,

AB^2

AC^2 + BC^2

⇒

c^2

a^2 + b^2

........ (i)
From ΔABC,
cosec B =

{AB}/{AC}

c/b

.... (ii)
cos A =

{AC}/{AB}

b/c

∴ cosec B - cos A =

c/b-b/c

{c^2-b^2}/{bc}

a^2/{bc}

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