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

Answered by puissant last updated on 19/Jul/21

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

Commented by nadovic last updated on 19/Jul/21

$$Niceone$$

Commented by mathdanisur last updated on 19/Jul/21

$$thankyouSircool$$

Commented by puissant last updated on 03/Sep/21

$$Q863$$

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