$(B)\,\,T$
$(C)\,\,T$
$(D)\,\, T$ $ \frac{1}{{\sqrt {1\,\, + \,\,{x^2}} }}$ ,
$(A)$ $y = tan(cos^{ -1} x)$ note that $0 < sin-1x < p$ but $tan(\pi /2)$ is not defined hence $cos^{ -1} x \,\neq$ $\frac{\pi }{2}\,$ $\Rightarrow x \neq 0$ hence domain is $[-1,1] - \{0\}$ and range is $R$. The graph is as shown.
$(B)$ $y = tan(cot^{ -1} x)$ note that $0 < cot-1x < \pi$ $\Rightarrow tan(\pi /2)$ is not defined hence $cos^{ -1} x \Rightarrow x 0$ hence domain $= R - \{0\}$ since $cot^{ -1} x \neq 0 , \pi \,\Rightarrow y \neq 0$ hence range is $R - \{0\}$ The graph is as shown.
$(C)$ $y = sin(tan^{ -1} x)$ $-\frac{\pi }{2}\,$ $< tan^{ -1} x <$ $\frac{\pi }{2}\,$ $\Rightarrow y (-1, 1)$ since $tan^{ -1} x$ is defined for all $x \in R$ and sin is also defined for all $x \in R$ $\Rightarrow$ domain is $R$ $y =$ $\frac{x}{{\sqrt {1 + {x^2}} }}\,$ The graph is as shown.
$(D)$ $y = cos(tan^{ -1} x)$ $tan^{ -1}$ $\, \in \,$ $\left( { - \,\frac{\pi }{2}\,,\,\frac{\pi }{2}} \right)$ $\Rightarrow y (0, 1]$ Domain is $R$ $y =$ $\frac{1}{{\sqrt {1 + {x^2}} }}$ Note that $sin^{ -1} (cot^{ -1} x)$ also has the same graph The graph is as shown.
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(where [.], {.} and $sgn\ x$ denotes greatest integer function, fractional part function and signum function respectively)
$x - cy - cz = 0 \,\,;\,\, cx - y + cz = 0 \,\,;\,\, cx + cy - z = 0 $ has a non -trivial solution, is