Mathematics

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

$89 / 2$

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

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

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

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

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

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

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

$=\dfrac{{\left(\tan{{89}^{\circ}}\right)}^{2013}+1}{1+{\left(\tan{{89}^{\circ}}\right)}^{2013}}+\dfrac{{\left(\tan{{88}^{\circ}}\right)}^{2013}+1}{1+{\left(\tan{{88}^{\circ}}\right)}^{2013}}+\dfrac{{\left(\tan{{87}^{\circ}}\right)}^{2013}+1}{1+{\left(\tan{{87}^{\circ}}\right)}^{2013}}+...up \,to \,44\, terms+{45}^{th}\,term$

$=1+1+1+...up \,to \,44\, terms+\dfrac{1}{1+{\left(\tan{{45}^{\circ}}\right)}^{2013}}$

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

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

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

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

$\therefore,\,\sum_{\theta={1}^{\circ}}^{{89}^{\circ}}{f\left(\theta\right)}=\dfrac{89}{2}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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