इन समीकरण को हल कीजिये `sec^(-1)""x/a-sec^(-1)""x/b=sec^(-1)b-sec^(-1)a`

Explanation

दिया गया समीकरण है -
`sec^(-1)((x)/(a))-sec^(-1)((x)/(b))=sec^(-1)(b)-sec^(-1)(a)`
`rArr cos^(-1)((a)/(x))-cos^(-1)((b)/(x))=cos^(-1)((1)/(b))-cos^(-1)((1)/(a))," "[because sec^(-1)x=cos^(-1)((1)/(x))]`
`rArr cos^(-1)((a)/(x))+cos^(-1)((1)/(a))=cos^(-1)((1)/(b))+cos^(-1)((b)/(x))`
`rArr cos^(-1)[(a)/(x).(1)/(a)-sqrt(1-(a^(2))/(x^(2)))sqrt(1-(1)/(a^(2)))]`
`" "=cos^(-1)[(1)/(b).(b)/(x)-sqrt(1-(1)/(b^(2)))sqrt(1-(b^(2))/(x^(2)))]`
`rArr cos^(-1)[(1)/(x)-sqrt((x^(2)-a^(2))/(x^(2)))sqrt((a^(2)-1)/(a^(2)))]`
`=cos^(-1)[(1)/(x)-sqrt((b^(2)-1)/(b^(2)))sqrt((x^(2)-b^(2))/(x^(2)))]`
`rArr (1)/(x)-sqrt((x^(2)-a^(2))/(x^(2)))sqrt((a^(2)-1)/(a^(2)))`
`rArr sqrt((x^(2)-a^(2))/(x^(2)))sqrt((a^(2)-1)/(a^(2)))=sqrt((b^(2)-1)/(b^(2)))sqrt((x^(2)-b^(2))/(b^(2)))`
`rArr ((x^(2)-a^(2))(a^(2)-1))/(a^(2)x^(2))=((x^(2)-b^(2))(b^(2)-1))/(b^(2)x^(2)),`