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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)]
x ?8x ?16x ? 0, x?0? ? 0, x?0? ? 2
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If roots of (x-p)(x-q) = c are a and b what will be the roots of (x-a)(x-b) = -c please explain. Solution) (x-p)(x-q)=c x2-(p+q)x-c=0 hence, a+b=p+q and a.b=pq-c
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a) Determine the distance traveled among t = 0 and t =∏/2 by a particle P(x, y) whose position at time t is given by Also check your result geometrically. (5) b) D
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am i going to get As
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