Bcg matrix assignment, marketing, Mathematics

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Sin[cot-1{cos(tan-1x)}], sin (cot -1 {cos (tan -1 x)}) tan -1 x = A  ...

sin (cot -1 {cos (tan -1 x)}) tan -1 x = A  => tan A =x sec A = √(1+x 2 ) ==>  cos A = 1/√(1+x 2 )    so   A =  cos -1 (1/√(1+x 2 )) sin (cot -1 {cos (tan -1 x)}) = s

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