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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)]
ABC is a right triangle right-angled at C and AC=√3 BC. Prove that ∠ABC=60 o . Ans: Tan B = AC/BC Tan B = √3 BC/BC Tan B =√3 ⇒ Tan B = Tan 60 ⇒ B = 60
steps to trace the cartesian curve
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Question 1: (a) Show that, for all sets A, B and C, (i) (A ∩ B) c = A c ∩B c . (ii) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). (iii) A - (B ∪ C) = (A - B) ∩ (A - C).
write CxD being sure to use appropriate brackets and find n(CxD)
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