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cos(sin^-1 (3/5)+ cot^-1(3/2)) = 6/(5sqr...

`cos(sin^-1 (3/5)+ cot^-1(3/2)) = 6/(5sqrt13)`

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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
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