The magnitude of the resultant of two forces P and Q acting at a point is (2m+1) `sqrt(P^2+Q^2)` When the angle between them is `alpha ` , and is (2m-1) `sqrt(P^2+Q^2)` when the angle is `((pi)/(2)-alpha)`. Prove that, `tan alpha=(m-1)/(m+1)`.

Explanation

When the angle is `alpha` , the magnitude of the resultant is ,
`R=sqrt(P^2+Q^2+PQcos alpha)`
Given, `R=(2m+1)sqrt(P^2+Q^2).` On comparison,
`P^2+Q^2+2PQcos alpha =(2m+1)(P^2+Q^2)`
` or, 2PQcos alpha =(P^2+Q^2){(2m+1)^2-1}`
`=(P^2+Q^2)(4m^2+4m+1-1)`
`PQcos alpha =2(P^2+Q^2)m(m+1)`
`or,PQ cos alpha=2(P^2+Q^2)m(m+1)`
`=2m(m+1)(P^2+Q^2)......(1)`
Again , when the angle is `((pi)/(2)-alpha)` , the magnitude of the resultant is,
`R.=sqrt(P^2+Q^2+2PQ cos ((pi)/(2)-alpha))=sqrt(P^2+Q^2+2PQsin alpha)`
Given , `R.=(2m-1)sqrt(P^2+Q^2)` . on comparison,
`P^2+Q^2+2PQsin alpha=(2m-1)^2(P^2+Q^2)`
`or, 2PQ sin alpha =(P^2+Q^2){(2m-1)-1}`
`=(P^2+Q^2)(4m^2-4m+1-1)`
`or,PQsin alpha =2(P^2+Q^2)m(m-1)=2m(m-1)(P^2+Q^2)........(2)`
Dividing (2) by (1) we have , `tan alpha =(m-1)/(m+1)*`