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?
A. 2
B. 5
C. 6
D. 3
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & \frac{1}{{1 + {{\cot }^2}\theta }} + \frac{3}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} + 2{\sin ^2}\theta \cr & \Rightarrow \frac{1}{{{{\operatorname{cosec} }^2}\theta }} + \frac{3}{{{{\sec }^2}\theta }} + 2{\sin ^2}\theta \cr & \Rightarrow {\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta + 2{\sin ^2}\theta \cr & \Rightarrow 3\left( {{{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Rightarrow 3\left( 1 \right) \cr & \Rightarrow 3 \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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