Question

The value of {\sin ^{ - 1}}\left\\{ {\cot \left( {{\sin }^{ - 1}\sqrt {\dfrac{{2 + \sqrt 3 }}{4}}} + {{\cos }^{ - 1}}\left( {\dfrac{{\sqrt{12}}}{4}} \right) + {{{\sec }^{ - 1}}\sqrt {2}} \right) } \right\\} =
A). $1$
B). $0$
C). $- 1$
D). $2$

Answer 1

Simplify the expressions given in the innermost brackets first. Make the square roots as simple as possible and in a way where the value of that trigonometric function can be determined. Consider all the terms that are inside the bracket separately, then add their values finally. Then simplify the first two trigonometric functions using the answers that were given before. The basic formulas used are:
\eqalign{ & \sin \theta = \dfrac{{opposite}}{{hypotenuse}} \cr & \cos \theta = \dfrac{{adjacent}}{{hypotenuse}} \cr & \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} \cr & \sec \theta = \dfrac{1}{{\cos \theta }} \cr & {\tan ^{ - 1}}\left( {2 - \sqrt 3 } \right) = {15^ \circ } \cr & {\cos ^{ - 1}}\dfrac{{\sqrt 3 }}{2} = {30^ \circ } \cr & {\cos ^{ - 1}}\dfrac{1}{{\sqrt 2 }} = {45^ \circ } \cr & \cot \left( {{{90}^ \circ }} \right) = 0 \cr & {\sin ^{ - 1}}\left( 0 \right) = 0 \cr}

Complete step-by-step solution:
Let us first consider the three trigonometric expressions inside of the brackets. That is,
$\left( {{{\sin }^{ - 1}}\sqrt {\dfrac{{2 + \sqrt 3 }}{4}} } \right) + {\cos ^{ - 1}}\left( {\dfrac{{\sqrt {12} }}{4}} \right) + {\sec ^{ - 1}}\sqrt 2$
We will now solve them separately and add the values in the end.
Let, $\left( {{{\sin }^{ - 1}}\sqrt {\dfrac{{2 + \sqrt 3 }}{4}} } \right) = \theta$
If, $\left( {{{\sin }^{ - 1}}\sqrt {\dfrac{{2 + \sqrt 3 }}{4}} } \right) = \theta$, then $\tan \theta = \sqrt {\dfrac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }}}$ $\because \tan \theta = \dfrac{{\sin \theta }}{{\sqrt {1 - {{\sin }^{^2}}\theta } }}$
Rationalizing by $2 + \sqrt 3$, we get
$\tan \theta = \sqrt {\dfrac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }}} \times \sqrt {\dfrac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }}}$
$\Rightarrow \tan \theta = \sqrt {\dfrac{1}{{{{\left( {2 + \sqrt 3 } \right)}^2}}}}$
$\Rightarrow \tan \theta = \dfrac{1}{{2 + \sqrt 3 }}$
By taking reciprocal,
$\tan \theta = 2 - \sqrt 3$
$\Rightarrow \theta = {15^ \circ }$
Now consider the second term, ${\cos ^{ - 1}}\left( {\dfrac{{\sqrt {12} }}{4}} \right)$
This can be written as,${\cos ^{ - 1}}\left( {\dfrac{{\sqrt {4 \times } 3}}{{2 \times 2}}} \right)$
$\Rightarrow {\cos ^{ - 1}}\dfrac{{\sqrt 3 }}{2}$
We know, ${\cos ^{ - 1}}\dfrac{{\sqrt 3 }}{2} = {30^ \circ }$
Now, let us consider the last term, which is ${\sec ^{ - 1}}\sqrt 2$
We know that $\sec \theta = \dfrac{1}{{\cos \theta }}$
$\Rightarrow {\sec ^{ - 1}}\sqrt 2 = {\cos ^{ - 1}}\dfrac{1}{{\sqrt 2 }}$
The value of ${\cos ^{ - 1}}\dfrac{1}{{\sqrt 2 }} = {45^ \circ }$
We have now got values of all the three expressions. Let us substitute in the question.
\eqalign{ & {\sin ^{ - 1}}\left\\{ {\cot \left( {{{15}^ \circ } + {{30}^ \circ } + {{45}^ \circ }} \right)} \right\\} \cr & \Rightarrow {\sin ^{ - 1}}\left\\{ {\cot \left( {{{90}^ \circ }} \right)} \right\\} \cr}
The value of $\cot {90^ \circ } = 0$
This leaves us with,
${\sin ^{ - 1}}\left( 0 \right)$
Which is also equal to $0$. $\because \sin \left( 0 \right) = 0$
The final answer is $0$
Therefore option (2) is the correct answer.

Note: The correct values are to be substituted while simplifying for sin and cos. Memorize the values of each trigonometric function along with the angles. Memorize the basic conversions of any trigonometric function to sin and cos. Be very careful while simplifying the square roots. While substituting a term for $\theta$ and substituting for it, use appropriate conversions. The options have integers, therefore take all the angles in degrees and not in radians.