Question

plz explain how is this inverse trigonometric function actually possible???​

Answer 1

Answer:

First of all let's see the identity used in the first step of your solution

sin 3A = 3sinA - $4sin^{3}A$

$1/sinx$ can also be written as $sin^{-1}x$

So $sin^{-1}$(3sinθ - 4sinθ) was changed to  $sin^{-1}$(sin3θ) using the identity mentioned above in the second step.

now we have $sin^{-1}$(sin3θ), and the 1/sinx × sin3θ will be written as 3θ as both the sine got cancelled.

And in the final step, θ was written as $sin^{-1}$ x only if it was mentioned somewhere in the question.

--------------------------------------------------------------------------------------------------------------

Hoping it was helpful to you. You may mark my answer as the brainliest if you feel the same.

It was a pleasure helping you.