sin[cot-1{cos(tan-1x)}]
Last Activity: 11 Years ago
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)]
Last Activity: 4 Years ago
Hello student
Let, sin (cot-1 {cos (tan -1x)}) = T
Let, tan-1 x = A or T = sin (cot-1 {cos (A)})
=> tan A = x; sec A = √(1+x2) ==> cos A = 1/√(1+x2)
T = sin (cot-1 {cos (A)}) = sin (cot-1 {(1/√(1+x2))})
Let, cot-1 {(1/√(1+x2))} = B or T = sin (B)
{(1/√(1+x2))} = cotB ==> cosec B = {(√[(2+x2)/(1+x2)])} ==> sin B = {(√[(1+x2)/(2+x2)]}
Hence, T = sin (B) = √[(1+x2)/(2+x2)]
Therefore, the answer is, T = √[(1+x2)/(2+x2)]
Dear StudentPlease see the solution in the attachment.I hope this answer will help you.Thanks & RegardsYash Chourasiya
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Last Activity: 2 Years ago