Give the value of
a.sin^-1(cos x) (Limit solution to an acute angle)
b.tan (sin^-1x)
c. tan[tan^-1x+1/x-1+tan^1x-1/x]
2006-12-03 23:09:54 · 3 answers · asked by GHz 1 in Science & Mathematics ➔ Mathematics
Hmm let me think, maybe asking your trigonometry professor is much better... :D
2006-12-03 23:17:59 · answer #1 · answered by Anonymous · 0⤊ 1⤋
a.sin^-1(cos x)
.sin^-1(sin(pi/2- x))
=pi/2-x
b.tan (sin^-1x)
put (sin^-1x)=y
x=siny
now
tan(sin^-1x)=tan y
=siny/cosy
=siny/(1-sin^2y)^1/2
=x(1-x^2)^1/2
c. tan[tan^-1x+1/x-1+tan^-1x-1/x]
= tan[tan^-1x-1+tan^-1x]
=tan[2tan^-1x-1]
check the question
2006-12-03 23:36:55 · answer #2 · answered by Dupinder jeet kaur k 2 · 0⤊ 0⤋
a.sin^-1cosx=sin^-1sin(pi/2-x)
=pi/2-x
i hope by sin^-1 you mean sin inverse
2006-12-03 23:20:53 · answer #3 · answered by raj 7 · 0⤊ 0⤋