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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)]
Series - The Basics That topic is infinite series. So just define what is an infinite series? Well, let's start with a sequence {a n } ∞ n=1 (note the n=1 is for convenie
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7(y + 3) - 2(x + 2) = 14, 4 (y - 2) + 3(x - 3) = 2 Ans: 7(y + 3) - 2 (x+ 2) = 14 --------- (1) 4(y- 2) + 3(x - 3) = 2 ----------(2) From (1) 7y +21 -
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[3+tan20+tan80]/tan20+tan80
let setM={X,2X,4X} for any numberX .if average (arthemetic mean)of the number in setM is 14.what is the value of X?
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what is equizilent to 2/5
Mensuration surface area
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