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Question Number 221686 by Tawa11 last updated on 09/Jun/25
Commented by AlagaIbile last updated on 09/Jun/25
Let tan x = y, cos x = (1/( (√(y^2 + 1)))) ⇒ cos^(2 ) [cos^(-1) (1/( (√(y^2 + 1))))] ⇒ [cos (cos^(-1) (1/( (√(y^2 + 1)))))]^2 ⇒ [(1/( (√(y^2 + 1))))]^2 f(y) = (1/(y^2 + 1)) f(x) = (1/(x^2 + 1))
$$\:{Let}\:\mathrm{tan}\:\boldsymbol{{x}}\:=\:\boldsymbol{{y}},\:\mathrm{cos}\:\boldsymbol{{x}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{y}}^{\mathrm{2}} \:+\:\mathrm{1}}} \\ $$$$\Rightarrow\:\mathrm{cos}^{\mathrm{2}\:} \left[\mathrm{cos}^{-\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{y}}^{\mathrm{2}} \:+\:\mathrm{1}}}\right] \\ $$$$\Rightarrow\:\left[\mathrm{cos}\:\left(\mathrm{cos}^{-\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{y}}^{\mathrm{2}} \:+\:\mathrm{1}}}\right)\right]^{\mathrm{2}} \\ $$$$\Rightarrow\:\left[\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{y}}^{\mathrm{2}} \:+\:\mathrm{1}}}\right]^{\mathrm{2}} \\ $$$$\:{f}\left({y}\right)\:=\:\frac{\mathrm{1}}{\boldsymbol{{y}}^{\mathrm{2}} \:+\:\mathrm{1}}\: \\ $$$$\:{f}\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\:\mathrm{1}} \\ $$
Commented by Tawa11 last updated on 09/Jun/25
Thanks sir. I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by mr W last updated on 09/Jun/25
f(tan x)=cos^2 x=((cos^2 x)/(cos^2 x+sin^2 x))=(1/(1+tan^2 x)) replace tan x with x we get f(x)=(1/(1+x^2 ))
$${f}\left(\mathrm{tan}\:{x}\right)=\mathrm{cos}^{\mathrm{2}} \:{x}=\frac{\mathrm{cos}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}+\mathrm{sin}^{\mathrm{2}} \:{x}}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}} \\ $$$${replace}\:\mathrm{tan}\:{x}\:{with}\:{x}\:{we}\:{get} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$
Commented by Tawa11 last updated on 09/Jun/25
Thanks sir. I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Commented by Tawa11 last updated on 09/Jun/25
Sir, they claim Q221707 is correct. Please verify sir. I don′t mean to stress you. But I want to know the errors in the question. Thanks sir.
$$\mathrm{Sir},\:\mathrm{they}\:\mathrm{claim}\:\mathrm{Q221707}\:\:\mathrm{is}\:\mathrm{correct}. \\ $$$$\mathrm{Please}\:\mathrm{verify}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{mean}\:\mathrm{to}\:\mathrm{stress}\:\mathrm{you}. \\ $$$$\mathrm{But}\:\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{know}\:\mathrm{the}\:\mathrm{errors}\:\mathrm{in}\:\mathrm{the}\:\mathrm{question}. \\ $$$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 09/Jun/25
then why don′t you let them solve the question?
$${then}\:{why}\:{don}'{t}\:{you}\:{let}\:{them}\:{solve}\:{the}\: \\ $$$${question}? \\ $$
Commented by Tawa11 last updated on 09/Jun/25
True sir. I just feel like you will have better approach sir.
$$\mathrm{True}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{just}\:\mathrm{feel}\:\mathrm{like}\:\mathrm{you}\:\mathrm{will}\:\mathrm{have}\:\mathrm{better} \\ $$$$\mathrm{approach}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 09/Jun/25
as i have said, it is not a clear question. when it is not a clear question to me, how can i solve it and how can i judge your solution? i think they don′t have a clue and can′t even draw a clear diagram to show what they want to mean. the drawings you have shown are contridictory to each other. what should the three circles mean and how are they defined? actually it is more a non−sense than a real question which deserves so much attention from us. actually i have chosen to ignore it. an advice: don′t spend too much of your time for such pseudo good questions!
$${as}\:{i}\:{have}\:{said},\:{it}\:{is}\:{not}\:{a}\:{clear}\: \\ $$$${question}.\:{when}\:{it}\:{is}\:{not}\:{a}\:{clear} \\ $$$${question}\:{to}\:{me},\:{how}\:{can}\:{i}\:{solve}\:{it} \\ $$$${and}\:{how}\:{can}\:{i}\:{judge}\:{your}\:{solution}? \\ $$$${i}\:{think}\:{they}\:{don}'{t}\:{have}\:{a}\:{clue}\:{and} \\ $$$${can}'{t}\:{even}\:{draw}\:{a}\:{clear}\:{diagram}\:{to}\: \\ $$$${show}\:{what}\:{they}\:{want}\:{to}\:{mean}.\:{the} \\ $$$${drawings}\:{you}\:{have}\:{shown}\:{are}\: \\ $$$${contridictory}\:{to}\:{each}\:{other}.\:{what} \\ $$$${should}\:{the}\:{three}\:{circles}\:{mean}\:{and} \\ $$$${how}\:{are}\:{they}\:{defined}?\:{actually} \\ $$$${it}\:{is}\:{more}\:{a}\:{non}−{sense}\:{than}\:{a} \\ $$$${real}\:{question}\:{which}\:{deserves}\:{so} \\ $$$${much}\:{attention}\:{from}\:{us}.\:{actually} \\ $$$${i}\:{have}\:{chosen}\:{to}\:{ignore}\:{it}. \\ $$$${an}\:{advice}:\:{don}'{t}\:{spend}\:{too}\:{much} \\ $$$${of}\:{your}\:{time}\:{for}\:{such}\:{pseudo}\:{good} \\ $$$${questions}! \\ $$
Commented by Tawa11 last updated on 09/Jun/25
Sir, please check the solution I posted.
$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{check}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{I}\:\mathrm{posted}. \\ $$
Commented by fantastic last updated on 09/Jun/25
true ����
Commented by Tawa11 last updated on 09/Jun/25
Thanks sir. I will ignore it sir
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{ignore}\:\mathrm{it}\:\mathrm{sir} \\ $$
Answered by wewji12 last updated on 09/Jun/25
tan(x)=u f(u)=cos^2 (tan^(−1) (u)) cos(tan^(−1) (u))=(1/( (√(u^2 +1)))) ∴ cos^2 (tan^(−1) (u))=(1/(u^2 +1)) ∴f(x)=(1/(x^2 +1)) heh EZ???
$$\mathrm{tan}\left({x}\right)={u} \\ $$$${f}\left({u}\right)=\mathrm{cos}^{\mathrm{2}} \left(\mathrm{tan}^{−\mathrm{1}} \left({u}\right)\right) \\ $$$$\mathrm{cos}\left(\mathrm{tan}^{−\mathrm{1}} \left({u}\right)\right)=\frac{\mathrm{1}}{\:\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\therefore\:\mathrm{cos}^{\mathrm{2}} \left(\mathrm{tan}^{−\mathrm{1}} \left({u}\right)\right)=\frac{\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\therefore{f}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{heh}\:\mathrm{EZ}??? \\ $$
Commented by Tawa11 last updated on 09/Jun/25
Thanks sir. I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by efronzo1 last updated on 10/Jun/25
f(tan x)= cos^2 x= (1/(sec^2 x))=(1/(tan^2 x+1)) f(x) = (1/(x^2 +1))
$$\:\mathrm{f}\left(\mathrm{tan}\:\mathrm{x}\right)=\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}=\:\frac{\mathrm{1}}{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}}=\frac{\mathrm{1}}{\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}+\mathrm{1}} \\ $$$$\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\: \\ $$
Commented by Tawa11 last updated on 10/Jun/25
Thanks sir. I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

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