sin (cot^-1 {cos (tan ^-1x)})

tan^-1 x = A  => tan A =x

sec A = √(1+x²) ==>  cos A = 1/√(1+x²)    so   A =  cos^-1(1/√(1+x²))

sin (cot^-1 {cos (tan ^-1x)}) = sin (cot^-1 {cos (cos^-1(1/√(1+x²))}) 

=sin (cot^-1 {(1/√(1+x²))})

if cot^-1 {(1/√(1+x²))} = B

{(1/√(1+x²))} = cotB  ==>  cosec B = {(√[(2+x²)/(1+x²)])}

sin B = {(√[(1+x²)/(2+x²)]} ==>  B  = sin ^-1 ({(√[(1+x²)/(2+x²)]})

sin {sin ^-1 ({(√[(1+x²)/(2+x²)]})} = √[(1+x²)/(2+x²)]

the answer is √[(1+x²)/(2+x²)]

 

Posted Date: 3/9/2013 6:52:41 AM | Location : United States

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