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If (cos alpha)/(cos A)+(sin alpha)/(sin A)+(sin beta)/(sin A)=1, where alpha and beta do not differ by an even multiple of pi, prove that (cos alpha cos beta)/(cos^(2)A)+(sin alpha sin beta)/(sin^(2)A)=

Answer»

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`-1`
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Solution :`alpha` and `beta` are the ROOTS of the equation `(cos theta)/(cos A)+(sin theta)/(sin A)=1`
`rArr ((cos theta)/(cos A))^(2)=(1-(sin theta)/(sin A))^(2)`
`rArr (1-sin^(2)theta)/(cos^(2)A)=1-2(sin theta)/(sin A)+((sin theta)/(sin A))^(2)`
`rArr (1)/(cos^(2)A)=1-2(sin theta)/(sin A)+(sin^(2)theta)/(sin^(2)A)+(sin^(2))/(cos^(2)A)`
`rArr -(1)/(cos^(2)A)-2(sin theta)/(sin A)+(sin^(2)theta)/(sin^(2)A cos^(2)A)=0`
`rArr -(sin^(2)A)/(cos^(2)A)-2(sin theta)/(sin A)+(sin^(2)theta)/(sin^(2)A cos^(2)A)=0`
`rArr -(sin^(2)A)/(cos^(2)A)-2(sin theta)/(sin A)+(sin^(2)theta)(sin^(2)A cos(2)A)=0`
`rArr ((sin theta)/(sin A))^(2)-2((sin theta)/(sin A))cos^(2) A - sin^(2) A = 0`
Clearly,
`(sin alpha)/(sin A)` and `(sin beta)` are the roots of this equation
`therefore (sin alpha)/(sin A)(sin B)/(sin A)=-sin^(2)A`
`rArr (sin alpha sin beta)/(sin^(2)A)=-sin^(2)A`
Similarly, by MAKING a quadratic in ......(1)
`(cos theta)/(cos A)`, we get
`(cos alpha cos beta)/(cos^(2)A)=-cos^(2)A`.....(2)
Adding (1) and (2), we get
`(cos alpha cos beta)/(cos^(2)A)+(sin alpha sin beta)/(sin^(2)A)=-(cos^(2)A+sin^(2)A)=-1`


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