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Question
If ∆ABC is a right triangle and if ∠A = `pi/2` then prove that cos B – cos C = `- 1 + 2sqrt(2) cos "B"/2 sin "C"/2`
Solution
A + B + C = 180°
Given A = 90°
∴ B + C = 90°
⇒ `("B" + "C")/2` = 45°
`"B"/2 + "C"/2` = 45°
R.H.S = `- 1 + 2sqrt(2) cos "B"/2 sin "C"/2`
= `1 + sqrt(2) (2cos "B"/2 sin "C"/2)`
We know that 2 cosA sinB = sin(A + B) – sin(A – B)
= `- 1 + sqrt(2) (sin (("B" + "C"))/2 - sin (("B" - "C"))/2)`
= `- 1 + sqrt(2) (sin 45^circ - sin (("B" - "C"))/2)`
= `- 1 + sqrt(2) (1/sqrt(2) - sin (("B" - "C")^2)/2)`
= `- 1 + 1 - sqrt(2) sin (("B" - "C"))/2`
= `- sqrt(2) sin (("B" - "C"))/2` .....(1)
L.H.S = cos B – cos C
= `2 sin (("B" + "C"))/2 sin (("C" - "B"))/2`
= `2 sin 45^circ sin (("C" - "B"))/2`
= `2(1/sqrt(2)) sin ((-("B" - "C"))/2)`
= `- sqrt(2) sin (("B" - "C"))/2` .....(2)
From (1) and (2)
⇒ L.H.S = R.H.S
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