Question

let $f\left(\theta \right)=\frac{1}{1+(\tan \theta {)}^{2013}}$ then value of ${\sum }_{\theta =1^0}^{{89}^{\circ }}f\left(\theta \right)$ equals

• A. 45
• B. 44
• C. $89/2$
• D. $91/2$

Answer 1

The correct option is C $89/2$

Given:$f\left(\theta \right)=\frac{1}{1+{\left(\tan \theta \right)}^{2013}}$

${\sum }_{\theta =1^{\circ }}^{{89}^{\circ }}f\left(\theta \right)$

$=f\left(1^{\circ }\right)+f\left(2^{\circ }\right)+f\left(3^{\circ }\right)+...+f\left({\theta }_{{89}^{\circ }}\right)$

$=\frac{1}{1+{\left(\tan 1^{\circ }\right)}^{2013}}+\frac{1}{1+{\left(\tan 2^{\circ }\right)}^{2013}}+\frac{1}{1+{\left(\tan 3^{\circ }\right)}^{2013}}+.....+\frac{1}{1+{\left(\tan {89}^{\circ }\right)}^{2013}}$

$=\frac{1}{1+{\left(\tan ({90}^{\circ }-{89}^{\circ })\right)}^{2013}}+\frac{1}{1+{\left(\tan ({90}^{\circ }-{88}^{\circ })\right)}^{2013}}+\frac{1}{1+{\left(\tan ({90}^{\circ }-{87}^{\circ })\right)}^{2013}}+.....+\frac{1}{1+{\left(\tan {89}^{\circ }\right)}^{2013}}$

$=\frac{1}{1+{\left(\cot {89}^{\circ }\right)}^{2013}}+\frac{1}{1+{\left(\cot {88}^{\circ }\right)}^{2013}}+\frac{1}{1+{\left(\cot {87}^{\circ }\right)}^{2013}}+....+\frac{1}{1+{\left(\tan {87}^{\circ }\right)}^{2013}}+\frac{1}{1+{\left(\tan {88}^{\circ }\right)}^{2013}}+\frac{1}{1+{\left(\tan {89}^{\circ }\right)}^{2013}}$

$=\frac{{\left(\tan {89}^{\circ }\right)}^{2013}}{1+{\left(\tan {89}^{\circ }\right)}^{2013}}+\frac{{\left(\tan {88}^{\circ }\right)}^{2013}}{1+{\left(\tan {88}^{\circ }\right)}^{2013}}+\frac{{\left(\tan {87}^{\circ }\right)}^{2013}}{1+{\left(\tan {87}^{\circ }\right)}^{2013}}+...+\frac{1}{1+{\left(\tan {87}^{\circ }\right)}^{2013}}+\frac{1}{1+{\left(\tan {88}^{\circ }\right)}^{2013}}+\frac{1}{1+{\left(\tan {89}^{\circ }\right)}^{2013}}$

$=\frac{{\left(\tan {89}^{\circ }\right)}^{2013}+1}{1+{\left(\tan {89}^{\circ }\right)}^{2013}}+\frac{{\left(\tan {88}^{\circ }\right)}^{2013}+1}{1+{\left(\tan {88}^{\circ }\right)}^{2013}}+\frac{{\left(\tan {87}^{\circ }\right)}^{2013}+1}{1+{\left(\tan {87}^{\circ }\right)}^{2013}}+...upto44terms+{45}^{th}term$

$=1+1+1+...upto44terms+\frac{1}{1+{\left(\tan {45}^{\circ }\right)}^{2013}}$

$=1\times 44+\frac{1}{1+1^{2013}}$

$=44+\frac{1}{1+1}$

$=44+\frac{1}{2}$

$=\frac{88+1}{2}=\frac{89}{2}$

$\therefore ,{\sum }_{\theta =1^{\circ }}^{{89}^{\circ }}f\left(\theta \right)=\frac{89}{2}$