माना `sin^(-1)""(3)/(5)=theta`
तब `sin theta=(3)/(5)implies tan theta=(3)/(4)`
`implies theta=tan^(-1)""(3)/(4)`
अब , माना
बायाँ पक्ष `=cos(sin^(-1)""(3)/(5)+cot^(-1)""(3)/(2))`
`= cos(theta+tan^(-1)""(2)/(3))`
` = cos(tan^(-1)""(3)/(4)+tan^(-1)""(2)/(3))`
`= cos[tan^(-1)(((3)/(4)+(2)/(3))/(1-(3)/(4)xx(2)/(3)))]`
`=cos(tan^(-1)""(17)/(6))=cos(cos^(-1)""(6)/(sqrt(325)))`
`= (6)/(sqrt(325))=(6)/(5sqrt(13))=` दायाँ पक्ष