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

A
$\sqrt{5}$
B
$\frac{\sqrt{5}}{3}$
C
1
D
$\frac{2}{3}$

Solution:

We have,

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

The value of $x$ where $x > 0$ and $\tan (\sec^{- 1} (\frac{1}{x})) = \sin (\tan^{- 1} 2)$ , is

Solution:

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