यदि `sin ^( -1) "" (2a ) /( 1 + a ^ 2) - cos ^ ( -1) "" ( 1 - b ^ 2 ) /( 1 + b ^ 2 ) = tan ^( -1) "" (2x ) / ( 1 - x ^ 2 ) ` हो, तो सिद्ध कीजिए कि
` x = ( a - b) / ( 1 + ab ) `

Explanation

यहाँ ` sin ^( -1) "" ( 2a ) /( 1 + a ^ 2 ) - cos ^( -1) "" ( 1 - b ^ 2 ) /( 1 + b ^ 2 ) = tan ^( -1) "" ( 2x ) /( 1 - x ^ 2 ) " " `... (1)
माना ` a = tan theta, b = tan phi , x = tan Psi `
समी. (1 ) में इन मानो को रखने पर,
` sin ^( -1) ((2 tan theta ) /( 1 + tan ^( 2) theta)) - cos ^( -1) (( 1 - tan ^ 2 phi ) /( 1 + tan ^2 phi )) = tan ^(-1) ((2 tan Psi ) /( 1- tan ^2 Psi ) ) `
` rArr sin ^( -1) ( sin 2 theta ) - cos ^( -1) ( cos 2 phi ) = tan ^( -1) ( tan 2 Psi ) `
` rArr 2 theta - 2 phi = 2 Psi `
` rArr theta - phi = Psi `
` rArr tan ^( -1) a - tan ^( -1) b = tan ^( -1) x `
` rArr tan ^( -1) (( a - b ) /( 1+ ab)) = tan^( -1) x `
` rArr x = ( a - b ) /( 1 + ab ) `