In a triangle `ABC, sin A- cos B = Cos C`, then angle `B` is

Options

A. `pi`
B. `pi//3`
C. `pi//2`
D. `pi//4`

Answer

C

Explanation

In a `Delta`ABC, we have
`sin A - cos B = cos C rArr sin A = cos B + cos C`
`rArr "2 sin"(A)/(2)."cos"(A)/(2)=2cos((B+C)/(2)).cos((B-C)/(2))" "[because sin 2 A = 2 sin A . cos A]`
`and cos B + cos C = 2cos((B+C)/(2).cos((B-C)/(2))`
`rArr "2 sin"(A)/(2)."cos"(A)/(2)=2cos(90^(@)-(A)/(2)).cos((B-C)/(2))" "[because A + B + C = 180^(@) rArr ((B+C)/(2))=90^(@)-(A)/(2)]`
`rArr "2 sin"(A)/(2)."cos"(A)/(2)="2 sin"(A)/(2).cos((B-C)/(2))" "[because cos(90^(@)-theta)=sin theta]`
`rArr "cos"(A)/(2)=cos((B-C)/(2))`
`rArr (A)/(2)=(B-C)/(2)`
`rArr A + C = B" "...(i)`
Also, `A + C = 180^(@)-B" "...(ii)`
So, `180^(@)-B=B`
`rArr 2 B = 180^(@)`
`therefore B = 90^(@)`