Question

plz help ill post the question just nw...

Answer 1

\[\sin^{-1} x -\cos^{-1} x=\Pi/6\]

Answer 2

solve for x

Answer 3

\(cos^- (x) = \dfrac{\pi}{2}-sin^- (x)\)
so that the expression turns to
\[sin^- -\dfrac{\pi}{2}+sin^-(x)=\dfrac {\pi}{6}\\2sin^-(x) =\dfrac{\pi}{6}+\dfrac{\pi}{2}=\dfrac{2\pi}{3} \\sin^-(x) = \dfrac{\pi}{3}\\x= \dfrac{\sqrt{3}}{2}\]

Answer 4

but in 1st step it is pi by 6 not 2

Answer 5

read the solution carefully, please

Answer 6

oh i get it (using identity sin-x + cos-x = pi/2

:) thnx a lot @ooops