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Question
Prove the following:
In any triangle ABC, sin A − cos B = cos C then ∠B = `pi/2`.
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Solution
sin A − cos B = cos C
∴ sin A = cos B + cos C
∴ `2sin "A"/2 cos "A"/2 = 2cos(("B" + "C")/2)cos (("B" - "C")/2)`
∴ `2sin "A"/2 cos "A"/2 = 2cos(pi/2 - "A"/2)cos(("B" - "C")/2) ...[(because "A" + "B" + "C" = pi","),(therefore ("B" + "C")/2 = pi/2 - "A"/2)]`
∴ `2sin "A"/2 cos "A"/2 = 2sin "A"/2cos (("B" - "C")/2)`
∴ `cos "A"/2 = cos (("B" - "C")/2) ....[because sin "A"/2 ≠ 0]`
∴ `"A"/2 = ("B" - "C")/2`
∴ A = B – C .…(i)
In ΔABC,
A + B + C = π
∴ B – C + B + C = π …[From (i)]
∴ 2B = π
∴ B = `pi/2`
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