Question

sin inverse [cos(sin inverse x) ]+cos inverse [sin (cos inverese x) ]

Answer 1

Given

$sin^{-1}${$cos^{}$[$sin^{-1}x$]} + $cos^{-1}${$sin^{}$[$cos^{-1}x$]}

but , $sin^{-1}x$ + $cos^{-1}x$ = $\pi/2$

so,

$sin^{-1}${$cos^{}$[ $\pi/2$-$cos^{-1}x$]} + $cos^{-1}${$sin^{}$[ $\pi/2$-$sin^{-1}x$]}

Apply cos(a-b) and sin(a-b) formula

{

hence,

$sin^{-1}${$cos^{}$($\pi/2$)$cos^{}$($cos^{-1}x$) + $sin^{}$($\pi/2$)$sin^{}$($cos^{-1}x$)}

+

$cos^{-1}${$sin^{}$($\pi/2$)$cos^{}$($sin^{-1}x$) - $cos^{}$($\pi/2$)$sin^{}$($sin^{-1}x$)

}

$sin^{-1}${($sin^{}$($cos^{-1}x$) + $cos^{-1}$($cos^{}$($sin^{-1}$)}

but , $sin^{-1}$($cos^{}$ y)= y

and, $cos^{-1}$($cos^{}$ z)=z

here we have y =$cos^{-1} x$ and z=$sin^{-1} x$

so,

$sin^{-1}x$ + $cos^{-1}x$

$\pi/2$

Hope It helps.

Cheers!!!