?
The value of, $${\text{sec}}\theta \left( {\frac{{1 + \sin \theta }}{{{\text{cos}}\theta }} + \frac{{{\text{cos}}\theta }}{{1 + \sin \theta }}} \right)$$ - $$2{\text{ta}}{{\text{n}}^2}\theta $$ is?
Answer & Solution
Correct Answer:
Option
C
$$\eqalign{
& {\text{sec}}\theta \left( {\frac{{1 + \sin \theta }}{{{\text{cos}}\theta }} + \frac{{{\text{cos}}\theta }}{{1 + \sin \theta }}} \right) - 2{\text{ta}}{{\text{n}}^2}\theta \cr
& {\bf{Shortcut method:}} \cr
& {\text{Take, }}\theta = {0^ \circ } \cr
& \Rightarrow {\text{sec }}{0^ \circ }\left( {\frac{{1 + \sin {0^ \circ }}}{{{\text{cos }}{0^ \circ }}} + \frac{{{\text{cos }}{0^ \circ }}}{{1 + \sin {0^ \circ }}}} \right) - 2{\text{ta}}{{\text{n}}^2}{0^ \circ } \cr
& \Rightarrow 1\left( {\frac{{1 + 0}}{1} + \frac{1}{{1 + 0}}} \right) - 0 \cr
& \Rightarrow 2 \cr} $$
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