Question

Find the value of the expression given below,
{{\sin }^{-1}}\left\\{ \cot \left\\{ {{\sin }^{-1}}\left( \sqrt{\dfrac{2-\sqrt{3}}{4}} \right)+{{\cos }^{-1}}\dfrac{\sqrt{12}}{4}+{{\sec }^{-1}}\sqrt{2} \right\\} \right\\}
A.$\dfrac{\pi }{4}$
B.$\dfrac{\pi }{6}$
C.0
D.$\dfrac{\pi }{2}$

Answer 1

Hint: Convert the term $\sqrt{\dfrac{2-\sqrt{3}}{4}}$ into $\dfrac{\sqrt{3}-1}{2\sqrt{2}}$ as $\sin 15{}^\circ =\left( \dfrac{\sqrt{3}-1}{2\sqrt{2}} \right)$, similarly do the required adjustments to get the standard angle, after finding all the angles add them and take the ‘cot’ of it and then take ${{\sin }^{-1}}$ to get the final answer.

Complete step-by-step answer:
To solve the expression given expression we will first write it down and assume it as ‘S’, therefore,
S={{\sin }^{-1}}\left\\{ \cot \left\\{ {{\sin }^{-1}}\left( \sqrt{\dfrac{2-\sqrt{3}}{4}} \right)+{{\cos }^{-1}}\dfrac{\sqrt{12}}{4}+{{\sec }^{-1}}\sqrt{2} \right\\} \right\\}……………………………………………… (1)
For simplicity we will solve three terms involved in ‘cot’ separately, therefore,
Assume ${{\sin }^{-1}}\left( \sqrt{\dfrac{2-\sqrt{3}}{4}} \right)$ as ‘a’ therefore,
$a={{\sin }^{-1}}\left( \sqrt{\dfrac{2-\sqrt{3}}{4}} \right)$
As we know that $\sqrt{4}=2$ therefore we will get,
$a={{\sin }^{-1}}\left( \dfrac{\sqrt{2-\sqrt{3}}}{2} \right)$
To solve this we have to do an adjustment of multiplying and dividing the angle by $\sqrt{2}$ therefore we will get,
$\therefore a={{\sin }^{-1}}\left( \dfrac{\sqrt{2}}{\sqrt{2}}\times \dfrac{\sqrt{2-\sqrt{3}}}{2} \right)$
If we simplify the above equation we will get,
$\therefore a={{\sin }^{-1}}\left( \dfrac{\sqrt{2\left( 2-\sqrt{3} \right)}}{2\sqrt{2}} \right)$
$\therefore a={{\sin }^{-1}}\left( \dfrac{\sqrt{4-2\sqrt{3}}}{2\sqrt{2}} \right)$
Above equation can also be written as,
$\therefore a={{\sin }^{-1}}\left( \dfrac{\sqrt{3-2\sqrt{3}+1}}{2\sqrt{2}} \right)$
$\therefore a={{\sin }^{-1}}\left( \dfrac{\sqrt{{{\left( \sqrt{3} \right)}^{2}}-2\sqrt{3}+{{1}^{2}}}}{2\sqrt{2}} \right)$
To proceed further in the solution we should know the formula given below,
Formula:
${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$
By using the above formula in ‘a’ we will get,

$\therefore a={{\sin }^{-1}}\left( \dfrac{\sqrt{{{\left( \sqrt{3}-1 \right)}^{2}}}}{2\sqrt{2}} \right)$
By cancelling the square and square root we will get,
$\therefore a={{\sin }^{-1}}\left( \dfrac{\sqrt{3}-1}{2\sqrt{2}} \right)$
As we know that the value of $\sin 15{}^\circ =\left( \dfrac{\sqrt{3}-1}{2\sqrt{2}} \right)$ therefore we will get,
$\therefore a=15{}^\circ$
$\therefore {{\sin }^{-1}}\left( \sqrt{\dfrac{2-\sqrt{3}}{4}} \right)=15{}^\circ$ ……………………………………………………………. (2)
Assume ${{\cos }^{-1}}\dfrac{\sqrt{12}}{4}$ as b, therefore,
$b={{\cos }^{-1}}\dfrac{\sqrt{12}}{4}$
Above equation can also be written as,
$b={{\cos }^{-1}}\dfrac{\sqrt{4\times 3}}{4}$
$\therefore b={{\cos }^{-1}}\dfrac{2\sqrt{3}}{4}$
$\therefore b={{\cos }^{-1}}\dfrac{\sqrt{3}}{2}$
As we all know the value of $\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}$ therefore we will get,
$\therefore b=30{}^\circ$
$\therefore {{\cos }^{-1}}\dfrac{\sqrt{12}}{4}=30{}^\circ$ ……………………………………………………….. (3)
Assume ${{\sec }^{-1}}\sqrt{2}$ as ‘c’ therefore we will get,
$\therefore c={{\sec }^{-1}}\sqrt{2}$
As we all know the value of $\sec 45{}^\circ =\sqrt{2}$ therefore we will get,
$\therefore c=45{}^\circ$
$\therefore {{\sec }^{-1}}\sqrt{2}=45{}^\circ$ …………………………………………………….. (4)
Now we will put the values of equation (2), equation (3) and equation (4) in equation (1) to get,
\therefore S={{\sin }^{-1}}\left\\{ \cot \left\\{ 15{}^\circ +30{}^\circ +45{}^\circ \right\\} \right\\}
If add all the angles in the above equation we will get,
\therefore S={{\sin }^{-1}}\left\\{ \cot 90{}^\circ \right\\}
As we all know that the value of $\cot 90{}^\circ$ is 0 , if we put this value in the above equation we will get,
$\therefore S={{\sin }^{-1}}(0)$
Also, we know that the value of $\sin 0{}^\circ =0\Rightarrow 0{}^\circ ={{\sin }^{-1}}(0)$ therefore we will get,
$\therefore S=0{}^\circ$
As we have all the options in radians therefore we will convert the degrees into radians by simply multiplying it by $\dfrac{\pi }{180}$ therefore we will get,
$\therefore S=0\times \dfrac{\pi }{180}=0$
As we have assumed the given expression as ‘S’ therefore we will get,
\therefore {{\sin }^{-1}}\left\\{ \cot \left\\{ {{\sin }^{-1}}\left( \sqrt{\dfrac{2-\sqrt{3}}{4}} \right)+{{\cos }^{-1}}\dfrac{\sqrt{12}}{4}+{{\sec }^{-1}}\sqrt{2} \right\\} \right\\}=0
Therefore the value of the expression {{\sin }^{-1}}\left\\{ \cot \left\\{ {{\sin }^{-1}}\left( \sqrt{\dfrac{2-\sqrt{3}}{4}} \right)+{{\cos }^{-1}}\dfrac{\sqrt{12}}{4}+{{\sec }^{-1}}\sqrt{2} \right\\} \right\\} is equal to ‘0’.
Therefore, the correct answer is option (c).

Note: Do not write the angles in radians, as they will increase your calculation and make the calculations difficult. If you use the angles in degrees then you can solve it easily and then you can convert the final answer into radians by simply multiplying it by $\dfrac{\pi }{180}$.