Question

dooooooooooo thissssssssdd​

Answer 1

Inverse Trigonometry :

Solution :

Let $\mathsf{sin{\theta} = x}$ --> ( i )

Now, L. H. S. = $\mathsf{3 sin^{-1}x}$

We know that,

$\mathsf{sin3{\theta} = 3sin{\theta} - 4sin^{2}{\theta}}$

$\mathsf{3{\theta} = sin^{-1}[3sin{\theta} - 4sin^{2}{\theta}]}$

$\mathsf{3sin^{-1}x = sin^{-1}[3x - 4x^{2}]}$ - ( From i )

Hence, Proved.