Question

The value of x, where $x > 0$and $\tan \left\{ {{{\sec }^{ - 1}}\left( {\dfrac{1}{x}} \right)} \right\} = \sin \left( {{{\tan }^{ - 1}}2} \right)$.
1. $\sqrt 5 $
2. $\dfrac{{\sqrt 5 }}{3}$
3. $1$
4. $\dfrac{2}{3}$

Answer 1

Hint: It is given that x>0. This question contains inverse trigonometric functions. There is a sec function in the LHS along with tan. Convert the sec into tan. In RHS, sin and tan functions are present. Convert the tan into sin. Simplify the expressions inside the brackets first. Then we can find the value of x by equating LHS and RHS as per the given problem.
Formulae used:
$\eqalign{
  & {\tan ^{ - 1}}x = {\sin ^{ - 1}}\left( {\dfrac{x}{{\sqrt {{x^2} + 1} }}} \right) \cr
  & {\sec ^{ - 1}}x = {\tan ^{ - 1}}\left( {\sqrt {{x^2} - 1} } \right) \cr
  & \tan \left( {{{\tan }^{ - 1}}x} \right) = x \cr
  & \sin \left( {{{\sin }^{ - 1}}x} \right) = x \cr} $

Complete step-by-step solution:
The given equation is $\tan \left\{ {{{\sec }^{ - 1}}\left( {\dfrac{1}{x}} \right)} \right\} = \sin \left( {{{\tan }^{ - 1}}2} \right)$
LHS is given as $\tan \left\{ {{{\sec }^{ - 1}}\left( {\dfrac{1}{x}} \right)} \right\}$
Using the formula, we can simplify $\left\{ {{{\sec }^{ - 1}}\left( {\dfrac{1}{x}} \right)} \right\}$
For $x \geqslant 1$,${\sec ^{ - 1}}x = {\tan ^{ - 1}}\left( {\sqrt {{x^2} - 1} } \right)$
Since x>0, we can write LHS as
$\eqalign{
  & \Rightarrow \tan \left( {{{\tan }^{ - 1}}\dfrac{{\sqrt {1 - {x^2}} }}{x}} \right) \cr
  & \Rightarrow \dfrac{{\sqrt {1 - {x^2}} }}{x} \cr} $
Let us keep the LHS like this.
Next, simplify the RHS
RHS is given as $\sin \left( {{{\tan }^{ - 1}}2} \right)$
This can be simplified using the inverse trigonometric identities as mentioned above
$\eqalign{
  & \Rightarrow \sin \left( {{{\sin }^{ - 1}}\dfrac{2}{{\sqrt {{2^2} + 1} }}} \right) \cr
  & \Rightarrow \sin \left( {{{\sin }^{ - 1}}\dfrac{2}{{\sqrt 5 }}} \right) \cr
  & = \dfrac{2}{{\sqrt 5 }} \cr} $
This is the value of RHS.
Now let us equate the LHS and RHS.
$\dfrac{{\sqrt {1 - {x^2}} }}{x} = \dfrac{2}{{\sqrt 5 }}$
Squaring on both sides to remove the roots,
$\eqalign{
  & \Rightarrow \dfrac{{1 - {x^2}}}{{{x^2}}} = \dfrac{4}{5} \cr
  & \Rightarrow \dfrac{1}{{{x^2}}} - 1 = \dfrac{4}{5} \cr
  & \Rightarrow \dfrac{1}{{{x^2}}} = \dfrac{4}{5} + 1 \cr
  & \Rightarrow \dfrac{1}{{{x^2}}} = \dfrac{9}{5} \cr
  & \Rightarrow {x^2} = \dfrac{5}{9} \cr
  & \Rightarrow x = \dfrac{{\sqrt 5 }}{3} \cr} $
Therefore, the value of x is $\dfrac{{\sqrt 5 }}{3}$
Hence option (2) is the correct answer.

Note: The correct values are to be substituted while simplifying. Memorize the values of each trigonometric function along with the angles. Memorize the basic formulae of any trigonometric function to sin and cos. Be very careful while simplifying the expressions under square roots. Substitute properly for the given inverse trigonometric functions.