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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)]
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Proof of: if f(x) > g(x) for a x b then a ∫ b f(x) dx > g(x). Because we get f(x) ≥ g(x) then we knows that f(x) - g(x) ≥ 0 on a ≤ x ≤ b and therefore by Prop
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120
(6x+9y) + (11x+13y)
d
prove that s is bounded?
How to solve this: log x(81) = 4
Series Solutions to Differential Equations Here now that we know how to illustrate function as power series we can now talk about at least some applications of series. There ar
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