Mathematics

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


ANSWER

$$89 / 2$$


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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