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Question-65929




Question Number 65929 by Masumsiddiqui399@gmail.com last updated on 06/Aug/19
Commented by kaivan.ahmadi last updated on 06/Aug/19
cot1=tan89  cot2=tan88  ⋮  cot44=tan46  since  tan((π/2)−α)=cotα    S=(tan^2 1+cot^2 1)+(tan^2 2+cot^2 2)+...+(tan^2 44+cot^2 44)+tan^2 45=
$${cot}\mathrm{1}={tan}\mathrm{89} \\ $$$${cot}\mathrm{2}={tan}\mathrm{88} \\ $$$$\vdots \\ $$$${cot}\mathrm{44}={tan}\mathrm{46} \\ $$$${since}\:\:{tan}\left(\frac{\pi}{\mathrm{2}}−\alpha\right)={cot}\alpha \\ $$$$ \\ $$$${S}=\left({tan}^{\mathrm{2}} \mathrm{1}+{cot}^{\mathrm{2}} \mathrm{1}\right)+\left({tan}^{\mathrm{2}} \mathrm{2}+{cot}^{\mathrm{2}} \mathrm{2}\right)+…+\left({tan}^{\mathrm{2}} \mathrm{44}+{cot}^{\mathrm{2}} \mathrm{44}\right)+{tan}^{\mathrm{2}} \mathrm{45}= \\ $$
Commented by Tony Lin last updated on 08/Aug/19
Answered by MJS last updated on 06/Aug/19
((15931)/3)=((89×179)/3)
$$\frac{\mathrm{15931}}{\mathrm{3}}=\frac{\mathrm{89}×\mathrm{179}}{\mathrm{3}} \\ $$