Question

Rewrite the expression as an algebraic expression in x:
tan(sin^-1x)

Answer 1

sin^-1 is the same as 1/sin so
we would have\[\tan(x)=\frac{\sin(x) }{ \cos(x) }\]
so tan*1/sin is \[\frac{ \sin(x) }{\cos(x)*\sin(x) }\]
which then simplifies to (because the sins cancel out)
\[\frac{ 1 }{ \cos(x) } = \sec(x) \]

Answer 2

but \[\sin ^{-1}(x) \not =1/\sin(x) \]
\[\sin ^{-1}(x)= \]
\[\sin(\theta)=x\]

Answer 3

whoops sorry I misread yours

Setting sin^-1(x)=t (|t|<π/2),
we get x=sint and 1/(tan^2(t))+1=1/sin^2(t).
In particular tan^2(t)=1/(1/sin^2(t)-1)
=1/(1/x^2-1)=x^2/(1-x^2).
Hence, we have tant=tan(sin^-1(x))=x/√(1-x^2).

i didnt feel like making pretty formulas anymore

Answer 4

can you explain how you did that? I don't understand.

Answer 5

We're dealing with trig, so drawing a right triangle is a good place to start

Answer 6

Now add in the angle theta

Answer 7

sin(theta) = opp/hyp

sin(theta) = x/1

so this means we can update the drawing to get