Question

cos[tan-1{sin(cot-1 x)}] = [(1 + x^2)/(1 - x^2)]^-1

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sin[cot-1{cos(tan-1x)}]

4 years ago

Answers : (2)

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)]