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The numerical value of $$\frac{1}{{1 + {{\cot }^2}\theta }}$$ + $$\frac{3}{{1 + {\text{ta}}{{\text{n}}^2}\theta }}$$ + $$2{\sin ^2}\theta $$ will be?
Answer & Solution
Correct Answer:
Option
D
$$\eqalign{
& \frac{1}{{1 + {{\cot }^2}\theta }} + \frac{3}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} + 2{\sin ^2}\theta \cr
& \Rightarrow \frac{1}{{{{\operatorname{cosec} }^2}\theta }} + \frac{3}{{{{\sec }^2}\theta }} + 2{\sin ^2}\theta \cr
& \Rightarrow {\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta + 2{\sin ^2}\theta \cr
& \Rightarrow 3\left( {{{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta } \right) \cr
& \Rightarrow 3\left( 1 \right) \cr
& \Rightarrow 3 \cr} $$
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