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प्रश्न
`sin^-1 4/5+2tan^-1 1/3=pi/2`
उत्तर
LHS = `sin^-1 4/5+2tan^-1 1/3`
`=sin^-1 4/5+tan^-1{(2xx1/3)/(1-(1/3)^2)}` `[because2tan^-1x=tan^-1{(2x)/(1-x^2)}`
`=sin^-1 4/5+tan^-1{(2/3)/(8/9)}`
`=sin^-1 4/5+tan^-1 3/4`
`=sin^-1 4/5+cos^-1 1/(sqrt(1+9/16)` `[becausetan^-1x=cos^-1 1/sqrt(1+x^2)]`
`=sin^-1 4/5+cos^-1 1/(5/4)`
`=sin^-1 4/5+cos^-1 4/5`
`=pi/2=`RHS
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