# How do you find the solutions to #sin^-1x=tan^-1x#?

##### 2 Answers

I got

#arcsin(0) = 0#

#arctan(0) = 0#

graph{arcsinx - arctanx [-3.08, 3.08, -1.54, 1.54]}

This is something where you can recall that **angles**. Therefore, let us choose to subtract

#sin(arcsinx - arctanx) = sin0#

Then, we know that

#=> sin(arcsinx)cos(arctanx) - cos(arcsinx)sin(arctanx) = sin0#

#xcos(arctanx) - cos(arcsinx)sin(arctanx) = 0#

Since

#color(green)(cos(arctanx) = 1/(sqrt(1+x^2)))#

Next,

#a^2 + x^2 = 1#

#a = sqrt(1 - x^2)#

That means:

#color(green)(cos(arcsinx)) = sqrt(1 - x^2)/1 = color(green)(sqrt(1 - x^2))#

Lastly,

#color(green)(sin(arctanx) = x/(sqrt(1 + x^2)))#

So, we can substitute these results to get:

#x/(sqrt(1+x^2)) - (xsqrt(1 - x^2))/(sqrt(1 + x^2)) = 0#

#(x - xsqrt(1 - x^2))/(sqrt(1 + x^2)) = 0#

Now let's solve for

#x - xsqrt(1 - x^2) = 0#

#x = xsqrt(1 - x^2)#

#1 = sqrt(1 - x^2)#

#1 = 1 - x^2#

#x^2 = 0 => color(blue)(x = 0)#

This actually happens to be at the inflection point of

graph{arcsinx - arctanx [-3.08, 3.08, -1.54, 1.54]}

First of all, notice that both

Apply function

As we know,

Therefore, our equation takes form

Before reducing by

Indeed it is:

So, we found one solution:

Now let's reduce our equation by

Apply function

So,