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In an acute angled `DeltaABC` , if tan (A+B-C) = 1 and sec (B+C-A)=2 , then find the value of cos (4B-3A) .

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We have ,
`tan(A+B-C)=1=tan45^(@)`
`rArrA+B-C=45^(@)` . . . (1)
Also `sec(B+C-A)=2sec60^(@)`
`rArr B+C-A=60^(@)` . . .(2)
Adding equations (1) and (2) we get
`2B=105^(@) rArr B=52.5^(@)` . . .(3)
Subtracting equation (1) from equation (2) , we get
`2C-2A=15^(@)rArrC-A=7.5^(@)` . . . (4)
we know that A+B+C `=180^(@)`
`rArrA+C=180^(@)-52.5^(@)` [from(3)]
`rArrA+C=127.5^(@)` . . .(5)
`rArr-A+C=7.5^(@)` . . . (4)
Adding equations (4) and (5) , we get
`2C=135^(@)rArrC=67.5^(@)`
`:.A=127.5^(@)-67.5^(@)=60^(@)`
`:. cos (4B-3A)=cos(4xx52.5^(@)-3xx60^(@))=cos(210^(@)-180^(@))=cos30^(@)`
`=(sqrt(3))/(2)`

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