The values of $x$ satisfying $\tan ({\sec ^{ - 1}}x) = \sin \left( {{{\cos }^{ - 1}}\dfrac{1}{{\sqrt 5 }}} \right)$ is
A) $ \pm \dfrac{{\sqrt 5 }}{3}$
B) $ \pm \dfrac{3}{{\sqrt 5 }}$
C) $ \pm \dfrac{{\sqrt 3 }}{5}$
D) $ \pm \dfrac{3}{5}$

Hint: n the above question we can see that we have trigonometric ratios. We will first draw a right angle triangle and then we simplify the equation according to that, Then we will try to simplify the trigonometric ratio in their simple form as
$\sec \theta = \dfrac{1}{{\cos \theta }}$
We will assume that
 ${\sec ^{ - 1}}x = \theta $
We can write this as
$\dfrac{1}{{\sec }}x = \theta $
By cross multiplication it gives us
$\sec \theta = x$ .

Complete step by step answer:
Here we have $\tan ({\sec ^{ - 1}}x) = \sin \left( {{{\cos }^{ - 1}}\dfrac{1}{{\sqrt 5 }}} \right)$
Let us first draw the right angle triangle:

In the above figure we have XYZ is a right angled triangle and
 $\angle XYZ = \theta $
Here p is the perpendicular, b stands for base and h is the hypotenuse of the triangle.
Let us first take the left hand side of the equation:
Now let us assume that ${\sec ^{ - 1}}x = \theta $ .
We can write it as $\sec \theta = x$ .
We know that
$\sec \theta = \dfrac{1}{{\cos \theta }}$ , so by putting this we can write
$\dfrac{1}{{\cos \theta }} = x \Rightarrow \cos \theta = \dfrac{1}{x}$
We know that the trigonometric ratio of the cosine function :
 $\cos \theta = \dfrac{b}{h}$
By comparing here we have
$b = 1,h = x$
Now by Pythagoras theorem we can calculate the perpendicular i.e.
$p = \sqrt {{h^2} - {b^2}} $
By putting the value it gives us
 $p = \sqrt {{x^2} - 1} $
We can write them now in tangent form i.e.
$\tan = \dfrac{p}{b}$
By substituting the values of base and perpendicular we have:
$\tan \theta = \dfrac{{\sqrt {{x^2} - 1} }}{1}$
From the above we can calculate the value of $\theta $ by transferring tan to the other side i.e.
$\theta = \dfrac{{\sqrt {{x^2} - 1} }}{1} \times \dfrac{1}{{\tan }}$
The above expression can also be written as
$\theta = {\tan ^{ - 1}}\left( {\sqrt {{x^2} - 1} } \right)$
From the above we have assumed that
${\sec ^{ - 1}}x = \theta $ .
Now we can put this value back in the original form of the equation i.e.
$\tan \left( {{{\tan }^{ - 1}}\sqrt {{x^2} - 1} } \right)$
We will now solve the right hand side of the equation i.e.
$\sin \left( {{{\cos }^{ - 1}}\dfrac{1}{{\sqrt 5 }}} \right)$
Let us assume that
 $\left( {{{\cos }^{ - 1}}\dfrac{1}{{\sqrt 5 }}} \right) = \theta $
On simplifying it gives us
 $\cos \theta = \dfrac{1}{{\sqrt 5 }}$
Now we know that
 $\cos \theta = \dfrac{b}{h}$
By comparing here we have
$b = 1,h = \sqrt 5 $
Now by Pythagoras theorem we can calculate the perpendicular i.e.
$p = \sqrt {{h^2} - {b^2}} $
By putting the value it gives us
$p = \sqrt {{{(\sqrt 5 )}^2} - {1^2}} $
On simplifying
$\sqrt {5 - 1} = \sqrt 4 $
It gives us the value
$p = 2$
We can write them now in sine form i.e.
$\sin = \dfrac{p}{h}$
By substituting the values of hypotenuse and perpendicular we have:
$\sin \theta = \dfrac{2}{{\sqrt 5 }}$
We can calculate the value of $\theta $ by transferring sin to the other side i.e.
$\theta = \dfrac{2}{{\sqrt 5 }} \times \dfrac{1}{{\sin }}$ And,
 $\dfrac{1}{{\sin }}$ can be written as
 ${\sin ^{ - 1}}$
So it gives us
$\theta = {\sin ^{ - 1}}\dfrac{2}{{\sqrt 5 }}$
Again we have assume here that
 $\left( {{{\cos }^{ - 1}}\dfrac{1}{{\sqrt 5 }}} \right) = \theta $
Now by putting the value of $\theta $ back in the equation we have
$\sin \left( {{{\sin }^{ - 1}}\dfrac{2}{{\sqrt 5 }}} \right)$
So the new equation here we have
  $\tan \left( {{{\tan }^{ - 1}}\sqrt {{x^2} - 1} } \right) = \sin \left( {{{\sin }^{ - 1}}\dfrac{2}{{\sqrt 5 }}} \right)$
We will apply the basic trigonometric identities such as $\sin ({\sin ^{ - 1}}x) = x$ and $\tan ({\tan ^{ - 1}}x) = x$
By applying these identities in the above we can write them as:
$\sqrt {{x^2} - 1} = \dfrac{2}{{\sqrt 5 }}$
We will square both the sides of the equation:

${\left( {\sqrt {{x^2} - 1} } \right)^2} = {\left( {\dfrac{2}{{\sqrt 5 }}} \right)^2}$
By squaring both the sides of the equation we have :
${x^2} - 1 = \dfrac{4}{5}$
We will cross multiply them and solve them:
$5\left( {{x^2} - 1} \right) = 4 \Rightarrow 5{x^2} - 5 = 4$
By grouping the similar term together:
$5{x^2} = 5 + 4 \Rightarrow 5{x^2} = 9$
It gives us
${x^2} = \dfrac{9}{5}$
We have to find the value of $x$ , so we have :
$x = \sqrt {\dfrac{9}{5}} \Rightarrow x = \pm \dfrac{3}{{\sqrt 5 }}$
Hence the correct option is option (B) $ \pm \dfrac{3}{{\sqrt 5 }}$.

Note:
We should note that in the above solution we have assumed
 $\left( {{{\cos }^{ - 1}}\dfrac{1}{{\sqrt 5 }}} \right) = \theta $
We can write
${\cos ^{ - 1}} = \dfrac{1}{{\cos }}$
So by putting this in the above expression we can write
$\dfrac{1}{{\cos }} \times \dfrac{1}{{\sqrt 5 }} = \theta $
By cross multiplication of cosine or by transferring cosine to the other side, we have:
$\cos \theta = \dfrac{1}{{\sqrt 5 }}$