निम्नलिखित समीकरण को हल कीजिए -
`tan^(-1).(1-x)/(1+x)=sin^(-1).(2x)/(1+x^(2)).`

Explanation

दिया गया समीकरण है -
`tan^(-1).(1-x)/(1+x)=sin^(-1).(2x)/(1+x^(2))`
`tan^(-1)1-tan^(-1)x=2tan^(-1)x`
`(pi)/(2)=2tan^(-1)x+tan^(-1)x`
`rArr" "3tan^(-1)x=(pi)/(4)`
`rArr" "3tan^(-1)x=(pi)/(4)`
`rArr" "tan^(-1)=(pi)/(12)`
`rArr" "x=tan.(pi)/(12)=tan.(180^(@))/(12)`
`rArr" "x=tan15^(@)=tan(45^(@)-30^(@))`
`rArr" "x=(tan45^(@)-tan30^(@))/(1+tan45^(@)tan30^(@))`
`rArr" "x=(1-(1)/(sqrt3))/(1+(1)/(sqrt3))=((sqrt3-1)/(sqrt3))/((sqrt3+1)/(sqrt3))`
`rArr" "x=(sqrt3-1)/(sqrt3+1)=((sqrt3-1)(sqrt3-1))/((sqrt3+1)(sqrt3-1))`
`rArr" "x=((sqrt3-1)^(2))/((sqrt3)^(2)-(1)^(2))=((sqrt3)^(2)+1-2sqrt3)/(3-1)`
`rArr" "x=(3+1-2sqrt3)/(3-1)=(4-2sqrt3)/(2)`
`rArr" "x=2-sqrt3.`