Question

The value of ${{\sin }^{-1}}\left[ \cot \left( {{\sin }^{-1}}\sqrt{\dfrac{2-\sqrt{3}}{4}}+{{\cos }^{-1}}\dfrac{\sqrt{12}}{4}+{{\sec }^{-1}}\sqrt{2} \right) \right]$ is?
(a) 0
(b) $\dfrac{\pi }{2}$
(c) $\dfrac{\pi }{3}$
(d) None of these

Answer 1

Hint: Simplify the argument of the inverse cosine function by using the prime factorization of $\sqrt{12}$ and cancelling the common factors with the denominator. Now, to find the values of ${{\cos }^{-1}}\dfrac{\sqrt{3}}{2}$ and ${{\sec }^{-1}}\sqrt{2}$ find the value of the angle for which the value of the cosine function is $\dfrac{\sqrt{3}}{2}$ in the range of angles $\left[ 0,\pi \right]$ and the value of the secant function is $\sqrt{2}$ in the range of angles $\left[ 0,\dfrac{\pi }{2} \right)\cup \left( \dfrac{\pi }{2},\pi \right]$ respectively. Substitute the value of that angle in the given expression. For the term $\left( {{\sin }^{-1}}\sqrt{\dfrac{2-\sqrt{3}}{4}} \right)$, find the value of $\sin {{15}^{\circ }}$ using the identity $\cos 2\theta =1-2{{\sin }^{2}}\theta $ and substituting the value of $\theta $ equal to ${{15}^{\circ }}$. Check if we get $\sin {{15}^{\circ }}=\sqrt{\dfrac{2-\sqrt{3}}{4}}$ and substitute this angle in terms of radian. Finally, add all the angles and find the value of the co – tangent of this resultant sum of angles and find the value of inverse sine function of this co – tangent value in the range of angles $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$.

Complete step by step answer:
Here we have been provided with the expression ${{\sin }^{-1}}\left[ \cot \left( {{\sin }^{-1}}\sqrt{\dfrac{2-\sqrt{3}}{4}}+{{\cos }^{-1}}\dfrac{\sqrt{12}}{4}+{{\sec }^{-1}}\sqrt{2} \right) \right]$ and we are asked to find its value. Let us assume this expression as E, so we have,
$\Rightarrow E={{\sin }^{-1}}\left[ \cot \left( {{\sin }^{-1}}\sqrt{\dfrac{2-\sqrt{3}}{4}}+{{\cos }^{-1}}\dfrac{\sqrt{12}}{4}+{{\sec }^{-1}}\sqrt{2} \right) \right]$
Now, first we need to find the values of the inverse functions present inside the small bracket. We have the functions ${{\sin }^{-1}}\sqrt{\dfrac{2-\sqrt{3}}{4}},{{\cos }^{-1}}\dfrac{\sqrt{12}}{4}$ and ${{\sec }^{-1}}\sqrt{2}$ inside the bracket. Let us find their values one by one.
(1) ${{\sin }^{-1}}\sqrt{\dfrac{2-\sqrt{3}}{4}}$ means the value of angle whose value of the sine function is $\sqrt{\dfrac{2-\sqrt{3}}{4}}$ in the range of angles $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$. Now, we know that $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ so using the trigonometric identity $\cos 2\theta =1-2{{\sin }^{2}}\theta $ we have,
$\begin{align}
  & \Rightarrow \cos \left( 2\times {{15}^{\circ }} \right)=1-2{{\sin }^{2}}{{15}^{\circ }} \\
 & \Rightarrow \cos \left( {{30}^{\circ }} \right)=1-2{{\sin }^{2}}{{15}^{\circ }} \\
 & \Rightarrow \dfrac{\sqrt{3}}{2}=1-2{{\sin }^{2}}{{15}^{\circ }} \\
 & \Rightarrow {{\sin }^{2}}{{15}^{\circ }}=\dfrac{1}{2}\left( 1-\dfrac{\sqrt{3}}{2} \right) \\
\end{align}$
Taking square root both the sides we get,
$\begin{align}
  & \Rightarrow \sin {{15}^{\circ }}=\sqrt{\dfrac{1}{2}\left( 1-\dfrac{\sqrt{3}}{2} \right)} \\
 & \Rightarrow \sin {{15}^{\circ }}=\sqrt{\dfrac{1}{2}\left( \dfrac{2-\sqrt{3}}{2} \right)} \\
 & \Rightarrow \sin {{15}^{\circ }}=\sqrt{\dfrac{2-\sqrt{3}}{4}} \\
\end{align}$
On converting 15 degrees into radian we get $\dfrac{\pi }{12}$ radians, so we have,
$\Rightarrow \sin \dfrac{\pi }{12}=\sqrt{\dfrac{2-\sqrt{3}}{4}}$
Taking sine inverse function both the sides and using the formula $\sin \left( {{\sin }^{-1}}\theta \right)=\theta $ for $\theta \in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ we get,
\[\begin{align}
  & \Rightarrow \dfrac{\pi }{12}={{\sin }^{-1}}\left( \sqrt{\dfrac{2-\sqrt{3}}{4}} \right) \\
 & \Rightarrow {{\sin }^{-1}}\left( \sqrt{\dfrac{2-\sqrt{3}}{4}} \right)=\dfrac{\pi }{12}.........\left( i \right) \\
\end{align}\]
(2) ${{\cos }^{-1}}\dfrac{\sqrt{12}}{4}$ means the value of angle whose value of the cosine function is $\dfrac{\sqrt{12}}{4}$ in the range of angles $\left[ 0,\pi \right]$. First let us simplify the argument of the inverse cosine function given, so we get,
$\begin{align}
  & \Rightarrow {{\cos }^{-1}}\dfrac{\sqrt{12}}{4}={{\cos }^{-1}}\dfrac{2\sqrt{3}}{4} \\
 & \Rightarrow {{\cos }^{-1}}\dfrac{\sqrt{12}}{4}={{\cos }^{-1}}\dfrac{\sqrt{3}}{2} \\
\end{align}$
We know that the value of $\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$ in the range of angles $\left[ 0,\pi \right]$ so we can write the above expression as:
$\begin{align}
  & \Rightarrow {{\cos }^{-1}}\dfrac{\sqrt{3}}{2}=\dfrac{\pi }{6} \\
 & \Rightarrow {{\cos }^{-1}}\dfrac{\sqrt{12}}{4}=\dfrac{\pi }{6}..........\left( ii \right) \\
\end{align}$
(3) ${{\sec }^{-1}}\sqrt{2}$ means the value of angle whose value of the secant function is $\sqrt{2}$ in the range of angles $\left[ 0,\dfrac{\pi }{2} \right)\cup \left( \dfrac{\pi }{2},\pi \right]$. We know that $\sec \dfrac{\pi }{4}=\sqrt{2}$ and the $\dfrac{\pi }{4}$ lies the range $\left[ 0,\dfrac{\pi }{2} \right)\cup \left( \dfrac{\pi }{2},\pi \right]$. So we get,
$\Rightarrow {{\sec }^{-1}}\sqrt{2}=\dfrac{\pi }{4}............\left( iii \right)$
Substituting the values from equations (i), (ii) and (iii) in expression E we get,
$\begin{align}
  & \Rightarrow E={{\sin }^{-1}}\left[ \cot \left( \dfrac{\pi }{12}+\dfrac{\pi }{6}+\dfrac{\pi }{4} \right) \right] \\
 & \Rightarrow E={{\sin }^{-1}}\left[ \cot \left( \dfrac{\pi }{2} \right) \right] \\
\end{align}$
We know that $\cot \dfrac{\pi }{2}=0$, so we get,
$\Rightarrow E={{\sin }^{-1}}\left[ 0 \right]$
Therefore, we need to find the value of the angle whose value of the sine function is 0 and the angle must lie in the range $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$. We know that $\sin 0=0$, so we get,
$\begin{align}
  & \Rightarrow E={{\sin }^{-1}}\left( \sin 0 \right) \\
 & \therefore E=0 \\
\end{align}$

So, the correct answer is “Option a”.

Note: Note that there are many angles for which the values of trigonometric functions repeat its value as they are periodic functions with a certain period. This is the reason that certain ranges of the angles of different inverse trigonometric functions are defined as their principal values so that we get only one answer. You must remember the range of angles for all the inverse trigonometric functions. Also, try to remember the values of all the trigonometric functions for the angles ${{15}^{\circ }},{{36}^{\circ }}$ and ${{53}^{\circ }}$.