Question Number 56732 by kelly33 last updated on 22/Mar/19
Commented by kelly33 last updated on 22/Mar/19
$${find}\:{all}\:{parameter}\:{values}\:{a} \\ $$$${with}\:{presentation}\:{equation} \\ $$$${has}\:{a}\:{solution}. \\ $$$$ \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Mar/19
![tan^2 (π+2x)×cot^2 (−x)=2a tan^2 2x×cot^2 x=2a (((2tanx)/(1−tan^2 x)))^2 ×cot^2 x=2a a=(1/2)×((4tan^2 x×cot^2 x)/((1−tan^2 x)^2 )) a=(2/((1−tan^2 x)^2 )) now when tanx→0 a→2 [at x=0^o ] when tanx →+∞ or tanx→−∞ then tan^2 x →∞ then a→0 so i[think a [0,2]](https://www.tinkutara.com/question/Q56734.png)
$${tan}^{\mathrm{2}} \left(\pi+\mathrm{2}{x}\right)×{cot}^{\mathrm{2}} \left(−{x}\right)=\mathrm{2}{a} \\ $$$${tan}^{\mathrm{2}} \mathrm{2}{x}×{cot}^{\mathrm{2}} {x}=\mathrm{2}{a} \\ $$$$\left(\frac{\mathrm{2}{tanx}}{\mathrm{1}−{tan}^{\mathrm{2}} {x}}\right)^{\mathrm{2}} ×{cot}^{\mathrm{2}} {x}=\mathrm{2}{a} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{4}{tan}^{\mathrm{2}} {x}×{cot}^{\mathrm{2}} {x}}{\left(\mathrm{1}−{tan}^{\mathrm{2}} {x}\right)^{\mathrm{2}} } \\ $$$${a}=\frac{\mathrm{2}}{\left(\mathrm{1}−{tan}^{\mathrm{2}} {x}\right)^{\mathrm{2}} } \\ $$$${now}\:{when}\:{tanx}\rightarrow\mathrm{0}\:\:\:{a}\rightarrow\mathrm{2}\:\left[{at}\:{x}=\mathrm{0}^{{o}} \right] \\ $$$${when}\:{tanx}\:\rightarrow+\infty\:\:\:{or}\:{tanx}\rightarrow−\infty \\ $$$${then}\:{tan}^{\mathrm{2}} {x}\:\rightarrow\infty \\ $$$${then}\:{a}\rightarrow\mathrm{0} \\ $$$${so}\:{i}\left[{think}\:\:{a}\:\left[\mathrm{0},\mathrm{2}\right]\right. \\ $$$$ \\ $$