Question

Let $f\left(\theta \right)=\frac{1}{1+(\cot \theta {)}^2}$ and $S=\sum _{{\theta =1}^o}^{{89}^o}f\left(\theta \right)$. Then the value of $\sqrt{2S-8}=$

• A. $8$
• B. $9$
• C. $10$
• D. $12$

Answer 1

The correct option is B $9$

$\cot 1°=\tan (90°-1°)=\tan 89°=\frac{1}{\cot 89°}$

${\cot }^{2}1°=\frac{1}{{\cot }^{2}89°}$

$\frac{1}{{\cot }^{2}1°+1}=\frac{{\cot }^{2}89°}{1+{\cot }^{2}89°}$

So, $\frac{1}{1+{\cot }^{2}1°}+\frac{1}{1+{\cot }^{2}89°}=1$

So, $S={\sum }_{\theta =1}^{89}f\left(\theta \right)=44+\frac{1}{1+{\cot }^{2}45°}=44+\frac{1}{2}=\frac{89}{2}$

$\sqrt{2S-8}=\sqrt{89-8}=\sqrt{81}=9$