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In any triangle ABC, `sinA -cosB=cosC`, then angle B isA. `pi/2`B. `pi/3`C. `pi/4`D. `pi/6`

Answer» Correct Answer - a
We have, `sinA-cosB=cosC`
`sinA=cosB+cosC`
`rArr 2sinA/2cosA/2=2cos(B+C)/2cos(B-C)/2`
`rArr 2sinA/c cosA/2=2cos(pi-A)/2 cos(B-C)/(2)` `therefore A+B+C=pi`
`rArr 2sinA/2 cosA/2=2sinA/2cos(B-C)/(2)`
`rArr cosA/2=cos(B-C)/(2)` or `A=B-C`, But A+B+C=`pi`
Therefore `2B=pi rArr B=pi/2`


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