Domain of sin^-1x & cos^-1x is [-1, 1]

\(\therefore\) -1 \(\leq\) x \(\leq\) 1

If x = -1 or 0 or 1

then {x} = 0

Therefore, \(\frac{1+[x]}{x}\) is not defined.

\(\therefore\) x \(\neq\) -1, x \(\neq\) 0, x \(\neq\) 1

If 0 < x  < 1 then [x] = 0 and 0 < {x} < 1

Then \(\frac{1+[x]}{x}=\frac1{x}>1\)

which is not possible

because domain of cos^-1x is [-1. 1].

\(\therefore\) -1 < x < 0 then [x] = -1 and 0 < {x} < 1

\(\therefore\) \(\frac{1+[x]}{\{x\}}=\frac{1+(-1)}{\{x\}}=\frac0{\{x\}}=0\)

\(\because\) -1 < x < 0 ⇒ |x| = -x

\(\therefore\) L.H.S. = sin^-1x + cos^-1x = sin^-1(-x) + cos^-1x

 = -sin^-1x + cos^-1x (\(\because\) sin^-1(-x) = -sin^-1x)

 = \(-(\frac{\pi}2-cos^{-1}x)+cos^{-1}x\)

= \(-\frac{\pi}2+2cos^{-1}x\) 

R.H.S. = cos^-1(\(\frac{1+[x]}{\{x\}}\)) = cos^-1(0) = \(\frac{\pi}2\) 

\(\because\) L.H.S. = R.H.S.

\(\therefore\) \(-\frac{\pi}2+2cos^{-1}x = \frac{\pi}2\)

⇒ 2 cos^-1x = \(\frac{\pi}2+\frac{\pi}2=\pi\)

⇒ cos^-1x = \(\frac{\pi}2\) ⇒ x = cos(\(\frac{\pi}2\)) = 0

But -1 < x < 0

Hence, for no value of x given equality satisfies.