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Question Number 147272 by mathdanisur last updated on 19/Jul/21

Answered by puissant last updated on 19/Jul/21

y=arctan(a) ⇒ cos(arctan(a))=cos(y)  1+tan^2 y=(1/(cos^2 y)) ⇒ cos^2 y=(1/(1+tan^2 y))  ⇒cos^2 (arctan(a))=(1/(1+tan^2 (arctan(a))))=(1/(1+a^2 ))  ⇒cos(arctan(a))=(1/( (√(1+a^2 ))))  ...   sin^2 (arctan(−a))=1−cos^2 (arctan(−a))  =1−(1/(1+(−a)^2 ))=1−(1/(1+a^2 ))=(a^2 /(1+a^2 ))  ⇒ sin(arctan(−a))=((∣a∣)/( (√(1+a^2 ))))  a>0 ⇒ sin(arctan(−a))=(a/( (√(1+a^2 ))))    ((cos(arctan(a)))/(sin(arctan(−a))))=(1/( (√(1+a^2 ))))×((√(1+a^2 ))/a)=(1/a)....

$${y}={arctan}\left({a}\right)\:\Rightarrow\:{cos}\left({arctan}\left({a}\right)\right)={cos}\left({y}\right) \\ $$$$\mathrm{1}+{tan}^{\mathrm{2}} {y}=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {y}}\:\Rightarrow\:{cos}^{\mathrm{2}} {y}=\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} {y}} \\ $$$$\Rightarrow{cos}^{\mathrm{2}} \left({arctan}\left({a}\right)\right)=\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} \left({arctan}\left({a}\right)\right)}=\frac{\mathrm{1}}{\mathrm{1}+{a}^{\mathrm{2}} } \\ $$$$\Rightarrow{cos}\left({arctan}\left({a}\right)\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }} \\ $$$$...\:\:\:{sin}^{\mathrm{2}} \left({arctan}\left(−{a}\right)\right)=\mathrm{1}−{cos}^{\mathrm{2}} \left({arctan}\left(−{a}\right)\right) \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\left(−{a}\right)^{\mathrm{2}} }=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{a}^{\mathrm{2}} }=\frac{{a}^{\mathrm{2}} }{\mathrm{1}+{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{sin}\left({arctan}\left(−{a}\right)\right)=\frac{\mid{a}\mid}{\:\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }} \\ $$$${a}>\mathrm{0}\:\Rightarrow\:{sin}\left({arctan}\left(−{a}\right)\right)=\frac{{a}}{\:\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }} \\ $$$$\:\:\frac{{cos}\left({arctan}\left({a}\right)\right)}{{sin}\left({arctan}\left(−{a}\right)\right)}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}×\frac{\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{{a}}=\frac{\mathrm{1}}{{a}}.... \\ $$

Commented by nadovic last updated on 19/Jul/21

Nice one

$${Nice}\:{one} \\ $$

Commented by mathdanisur last updated on 19/Jul/21

thank you Sir cool

$${thank}\:{you}\:{Sir}\:{cool} \\ $$

Commented by puissant last updated on 03/Sep/21

Q863

$${Q}\mathrm{863} \\ $$

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