Show that? : # tan(arcsinx) = x/sqrt(1 -x^2) #

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Answer 1 · 2017-05-02T02:16:37

Let

# y=arcsinx iff x=siny #

Then using #sin^2A+cos^2A -= 1#; we have:

# sin^2y+cos^2y = 1 => x^2+cos^2y=1 #
# :. cos^2y = 1 -x^2 #
# :. sec^2y = 1/(1 -x^2) #

And, using the trig identity #tan^2A+1-=sec^2A#; we have:

# tan^2y+1-=sec^2y #

# :. tan^2y = sec^2y - 1#
# " " = 1/(1 -x^2) - 1#
# " " = (1-(1-x^2))/(1 -x^2) #
# " " = (x^2)/(1 -x^2) #

And so:

# :. tany = sqrt((x^2)/(1 -x^2)) #
# " " = x/sqrt(1 -x^2) #

But #y=arcsinx#; therefore:

# tan(arcsinx) = x/sqrt(1 -x^2) \ \ \ # QED