Question Number 90739 by ajfour last updated on 25/Apr/20
![solve for x and y tan^2 [π(x+y)]+cot^2 [π(x+y)] =1+(√((2x)/(x^2 +1)))](Q90739.png)
$${solve}\:{for}\:{x}\:{and}\:{y} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} \left[\pi\left({x}+{y}\right)\right]+\mathrm{cot}\:^{\mathrm{2}} \left[\pi\left({x}+{y}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}+\sqrt{\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$
Commented by me2love2math last updated on 26/Apr/20
$${max}\:{value}\:{or}\:{minimum}? \\ $$$$ \\ $$
Commented by ajfour last updated on 26/Apr/20
$${general}\:{values}. \\ $$
Answered by mr W last updated on 26/Apr/20
![tan^2 [π(x+y)]+cot^2 [π(x+y)]=1+(√((2x)/(x^2 +1))) tan^2 [π(x+y)]+cot^2 [π(x+y)]−2=(√((2x)/(x^2 +1)))−1 {tan [π(x+y)]−cot [π(x+y)]}^2 =(√((2x)/(x^2 +1)))−1 LHS={...}^2 ≥0 RHS=(√((2x)/(x^2 +1)))−1≤0 ⇒LHS=RHS=0 (√((2x)/(x^2 +1)))−1=0 ⇒x=1 tan [π(x+y)]−cot [π(x+y)]=0 tan^2 [π(x+y)]=1 tan [π(x+y)]=±1 ⇒π(x+y)=kπ±(π/4) ⇒x+y=k±(1/4) ⇒y=k−1±(1/4) ⇒y=n±(1/4) with n=any integer or ⇒y=(n/2)+(1/4) with n=any integer](Q90798.png)
$$\mathrm{tan}\:^{\mathrm{2}} \left[\pi\left({x}+{y}\right)\right]+\mathrm{cot}\:^{\mathrm{2}} \left[\pi\left({x}+{y}\right)\right]=\mathrm{1}+\sqrt{\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} \left[\pi\left({x}+{y}\right)\right]+\mathrm{cot}\:^{\mathrm{2}} \left[\pi\left({x}+{y}\right)\right]−\mathrm{2}=\sqrt{\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{1} \\ $$$$\left\{\mathrm{tan}\:\left[\pi\left({x}+{y}\right)\right]−\mathrm{cot}\:\left[\pi\left({x}+{y}\right)\right]\right\}^{\mathrm{2}} =\sqrt{\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{1} \\ $$$${LHS}=\left\{...\right\}^{\mathrm{2}} \:\geqslant\mathrm{0} \\ $$$${RHS}=\sqrt{\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{1}\leqslant\mathrm{0} \\ $$$$\Rightarrow{LHS}={RHS}=\mathrm{0} \\ $$$$\sqrt{\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{1} \\ $$$$\mathrm{tan}\:\left[\pi\left({x}+{y}\right)\right]−\mathrm{cot}\:\left[\pi\left({x}+{y}\right)\right]=\mathrm{0} \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\left[\pi\left({x}+{y}\right)\right]=\mathrm{1} \\ $$$$\mathrm{tan}\:\left[\pi\left({x}+{y}\right)\right]=\pm\mathrm{1} \\ $$$$\Rightarrow\pi\left({x}+{y}\right)={k}\pi\pm\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow{x}+{y}={k}\pm\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{y}={k}−\mathrm{1}\pm\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{y}={n}\pm\frac{\mathrm{1}}{\mathrm{4}}\:{with}\:{n}={any}\:{integer} \\ $$$${or} \\ $$$$\Rightarrow{y}=\frac{{n}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\:{with}\:{n}={any}\:{integer} \\ $$
Commented by mr W last updated on 26/Apr/20
$${a}\:{tricky}\:{question}!\:{i}\:{didn}'{t}\:{know}\:{how} \\ $$$${to}\:{solve}\:{straight}\:{away}. \\ $$
Commented by ajfour last updated on 26/Apr/20
$${Sir}\:{answer}\:{is}\:\:{x}=\mathrm{1},\:{y}=\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{4}}. \\ $$
Commented by mr W last updated on 26/Apr/20
$${y}={n}\pm\frac{\mathrm{1}}{\mathrm{4}}\:{and}\:{y}=\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{4}}\:{are}\:{identical}. \\ $$
Commented by ajfour last updated on 26/Apr/20
$${i}\:{did}\:{not}\:{doubt},\:{but}\:{couldn}'{t} \\ $$$${understand}\:{very}\:{well},\:{sir}. \\ $$$${Anyway}\:{thanks},\:{great}\:{solving}! \\ $$