If $f(x) = \frac{1 - x}{1 + x},$ then $f[f(\cos 2\theta)] = $

For every real number $x \neq -1$, let $f(x) = \frac{x}{x+1}$. Define $f_1(x) = f(x)$ and for $n \geq 2$, $f_n(x) = f(f_{n-1}(x))$. Then the product $f_1(-2) \cdot f_2(-2) \cdot \ldots \cdot f_n(-2)$ is equal to:

If $f(x) = \frac{1+x}{1-x}$ where $x \neq 1$,then $f(x) \cdot f(y) = $ . . . . . . .

If $f(x) = \frac{x}{\sqrt{1 + x^2}}$,then $(fofof)(x) = $

Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x)=2x+1$ and $g(x)=x^2-2$. Determine $(g \circ f)(x)$.

Let $f(x) = \sin x$ and $g(x) = \ln |x|$. If the ranges of the composite functions $fog$ and $gof$ are $R_1$ and $R_2$ respectively,then: