MCQ
If $f(x) = \frac{{1 - x}}{{1 + x}},$ then $f[f(\cos \;2\theta )] = $
- A$\tan 2\theta $
- B$\sec 2\theta $
- ✓$\cos 2\theta $
- D$\cot 2\theta $
$ = f({\tan ^2}\theta ) = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \cos \,\,2\theta .$
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$\left| {\begin{array}{*{20}{c}}a&{a + 1}&{a - 1}\\{ - b}&{b + 1}&{b - 1}\\c&{c - 1}&{c + 1}\end{array}} \right| + \left| {\begin{array}{*{20}{c}}{a + 1}&{b + 1}&{c - 1}\\{a - 1}&{b - 1}&{c + 1}\\{{{\left( { - 1} \right)}^{n + 2}} \cdot a}&{{{\left( { - 1} \right)}^{n + 1}} \cdot b}&{{{\left( { - 1} \right)}^n} \cdot c}\end{array}} \right| = 0$ then $n$ equals to