The value of x where x >0 and tan⁡sec−1⁡1x=sin⁡tan−1⁡2, is

# The value of $x$ where  and $\mathrm{tan}\left({\mathrm{sec}}^{-1}\left(\frac{1}{x}\right)\right)=\mathrm{sin}\left({\mathrm{tan}}^{-1}2\right)$, is

1. A

$\sqrt{5}$

2. B

$\frac{\sqrt{5}}{3}$

3. C

1

4. D

$\frac{2}{3}$

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### Solution:

We have,

$\begin{array}{l}\mathrm{tan}\left({\mathrm{sec}}^{-1}\left(\frac{1}{x}\right)\right)=\mathrm{sin}\left({\mathrm{tan}}^{-1}2\right)\\ ⇒\mathrm{tan}\left({\mathrm{tan}}^{-1}\frac{\sqrt{1-{x}^{2}}}{x}\right)=\mathrm{sin}\left({\mathrm{sin}}^{-1}\frac{2}{\sqrt{5}}\right)\\ ⇒\frac{\sqrt{1-{x}^{2}}}{x}=\frac{2}{\sqrt{5}}⇒\frac{1}{{x}^{2}}-1=\frac{4}{5}\\ ⇒\frac{1}{{x}^{2}}=\frac{9}{5}⇒x=\frac{\sqrt{5}}{3}\end{array}$

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