What is the value of $$\sin \left( {180 - \theta } \right)\sin \left( {90 - \theta } \right) + \frac{{\cot \left( {90 - \theta } \right)}}{{1 + {{\tan }^2}\theta }}?$$
A. 2cosθsinθ
B. $$\frac{{\cot \theta }}{{{{\left( {1 + {{\cot }^2}\theta } \right)}^2}}}$$
C. $$\frac{{\tan \theta }}{{{{\left( {1 + {{\tan }^2}\theta } \right)}^2}}}$$
D. sin2θcosθ
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & \sin \left( {180 - \theta } \right)\sin \left( {90 - \theta } \right) + \frac{{\cot \left( {90 - \theta } \right)}}{{1 + {{\tan }^2}\theta }} \cr & \Rightarrow \sin \theta \cos \theta + \frac{{\tan \theta }}{{{{\sec }^2}\theta }} \cr & \Rightarrow \sin \theta \cos \theta + \frac{{\sin \theta }}{{\cos \theta }} \times {\cos ^2}\theta \cr & \Rightarrow \sin \theta \cos \theta + \sin \theta \cos \theta \cr & \Rightarrow 2\cos \theta \sin \theta \cr} $$This Question Belongs to Arithmetic Ability >> Trigonometry
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