The numerical value of $$\frac{5}{{{\text{se}}{{\text{c}}^2}\theta }}$$ + $$\frac{2}{{1 + {\text{co}}{{\text{t}}^2}\theta }}$$ + $${\text{3}}{\sin ^2}\theta $$ is?
A. 5
B. 2
C. 3
D. 4
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & \frac{5}{{{\text{se}}{{\text{c}}^2}\theta }}{\text{ + }}\frac{2}{{1 + {\text{co}}{{\text{t}}^2}\theta }}{\text{ + 3}}{\sin ^2}\theta \cr & = 5{\cos ^2}\theta + \frac{2}{{{\text{cose}}{{\text{c}}^2}\theta }} + 3{\sin ^2}\theta \cr & = 5{\text{co}}{{\text{s}}^2}\theta + 2{\sin ^2}\theta + 3{\sin ^2}\theta \cr & = 5\left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right) \cr & \left( {\because {{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta = 1} \right) \cr & = 5 \times 1 \cr & = 5 \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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