Question

The value of \[{\cos ^{ - 1}}\left( {\sqrt {\dfrac{2}{3}} } \right) - {\cos ^{ - 1}}\left( {\dfrac{{\sqrt 6 + 1}}{{2\sqrt 3 }}} \right)\] is equal to
A. \[\dfrac{\pi }{3}\]
B. \[\dfrac{\pi }{4}\]
C. \[\dfrac{\pi }{2}\]
D. \[\dfrac{\pi }{6}\]

Answer 1

Hint: We use the formula \[{\cos ^{ - 1}}x \pm {\cos ^{ - 1}}y = {\cos ^{ - 1}}\left( {xy \pm \sqrt {1 - {x^2}} \sqrt {1 - {y^2}} } \right)\] and comparing with the given problem we will have x and y value. After simplifying we will get the required answer. We also use the formula \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] while simplifying. We know that square root and square will cancel out.

Complete step-by-step answer:
Given, \[{\cos ^{ - 1}}\left( {\sqrt {\dfrac{2}{3}} } \right) - {\cos ^{ - 1}}\left( {\dfrac{{\sqrt 6 + 1}}{{2\sqrt 3 }}} \right)\] . We have negative sign in-between cosine inverse.
So we have, \[{\cos ^{ - 1}}x - {\cos ^{ - 1}}y = {\cos ^{ - 1}}\left( {xy - \sqrt {1 - {x^2}} \sqrt {1 - {y^2}} } \right)\]
Comparing with the given problem, we have \[x = \sqrt {\dfrac{2}{3}} \] and \[y = \dfrac{{\sqrt 6 + 1}}{{2\sqrt 3 }}\]
Then,
 \[ = {\cos ^{ - 1}}\left( {\left( {\sqrt {\dfrac{2}{3}} \times \dfrac{{\sqrt 6 + 1}}{{2\sqrt 3 }}} \right) - \sqrt {1 - {{\left( {\sqrt {\dfrac{2}{3}} } \right)}^2}} \sqrt {1 - {{\left( {\dfrac{{\sqrt 6 + 1}}{{2\sqrt 3 }}} \right)}^2}} } \right)\] ------- (1)
We solve the terms \[\left( {\sqrt {\dfrac{2}{3}} \times \dfrac{{\sqrt 6 + 1}}{{2\sqrt 3 }}} \right)\] and \[\sqrt {1 - {{\left( {\sqrt {\dfrac{2}{3}} } \right)}^2}} \sqrt {1 - {{\left( {\dfrac{{\sqrt 6 + 1}}{{2\sqrt 3 }}} \right)}^2}} \] separately to avoid mistakes (Missing of terms), then substituting in the equation (1).
Now take, \[\left( {\sqrt {\dfrac{2}{3}} \times \dfrac{{\sqrt 6 + 1}}{{2\sqrt 3 }}} \right)\]
 \[ = \dfrac{{\sqrt 2 }}{{\sqrt 3 }} \times \dfrac{{\sqrt 6 + 1}}{{2\sqrt 3 }}\]
Using simple multiplication, we get:
 \[ = \dfrac{{\sqrt 2 \times (\sqrt 6 + 1)}}{{2 \times 3}}\]
Expanding the brackets in the numerator,
 \[ = \dfrac{{\sqrt {2 \times 6} + \sqrt 2 }}{6}\]
 \[ = \dfrac{{\sqrt {12} + \sqrt 2 }}{6}\]
We can again 12 has \[12 = 4 \times 3\]
 \[ = \dfrac{{\sqrt {4 \times 3} + \sqrt 2 }}{6}\]
We know the square root of 4 is 2.
 \[ = \dfrac{{2\sqrt 3 + \sqrt 2 }}{6}\] ------- (2).
Now, take \[\sqrt {1 - {{\left( {\sqrt {\dfrac{2}{3}} } \right)}^2}} \sqrt {1 - {{\left( {\dfrac{{\sqrt 6 + 1}}{{2\sqrt 3 }}} \right)}^2}} \]
We know that square and square root will cancel out, we get:
 \[ = \sqrt {1 - \dfrac{2}{3}} \sqrt {1 - \dfrac{{{{\left( {\sqrt 6 + 1} \right)}^2}}}{{{{\left( {2\sqrt 3 } \right)}^2}}}} \]
We use \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] ,
 \[ = \sqrt {\dfrac{{3 - 2}}{3}} \sqrt {1 - \dfrac{{{{\left( {\sqrt 6 } \right)}^2} + {1^2} + 2\sqrt 6 }}{{4 \times 3}}} \]
 \[ = \dfrac{1}{{\sqrt 3 }}\sqrt {1 - \dfrac{{(6 + 1 + 2\sqrt 6 )}}{{12}}} \]
Taking L.C.M. and simplifying,
 \[ = \dfrac{1}{{\sqrt 3 }}\sqrt {\dfrac{{12 - 7 - 2\sqrt 6 }}{{12}}} \]
 \[ = \dfrac{1}{{\sqrt 3 }}\sqrt {\dfrac{{5 - 2\sqrt 6 }}{{12}}} \]
 \[ = \dfrac{{\sqrt {(5 - 2\sqrt 6 )} }}{6}\]
We will simplify it further so that square root will get cancel, that is \[{\left( {\sqrt 2 - \sqrt 3 } \right)^2} = 2 + 3 - 2\sqrt 2 \sqrt 6 = 5 - 2\sqrt 6 \]
Substituting in above we get,
 \[ = \dfrac{{\sqrt {{{\left( {\sqrt 2 - \sqrt 3 } \right)}^2}} }}{6}\]
 \[ = \dfrac{{\left( {\sqrt 2 - \sqrt 3 } \right)}}{6}\] ----- (3).
Now substituting (2) and (3) in the equation (1) we get:
 \[ = {\cos ^{ - 1}}\left( {\dfrac{{2\sqrt 3 + \sqrt 2 }}{6} - \dfrac{{\left( {\sqrt 2 - \sqrt 3 } \right)}}{6}} \right)\]
Taking L.C.M and simplifying we get,
 \[ = {\cos ^{ - 1}}\left( {\dfrac{{2\sqrt 3 + \sqrt 2 - \sqrt 2 + \sqrt 3 }}{6}} \right)\]
Cancelling \[\sqrt 2 \] we get,
 \[ = {\cos ^{ - 1}}\left( {\dfrac{{2\sqrt 3 + \sqrt 3 }}{6}} \right)\]
Taking \[\sqrt 3 \] as common, we get,
 \[ = {\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 (2 + 1)}}{6}} \right)\]
 \[ = {\cos ^{ - 1}}\left( {\dfrac{{3\sqrt 3 }}{6}} \right)\]
Cancelling, we get:
 \[ = {\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)\]
We know that \[{\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = \dfrac{\pi }{6}\] , so we get:
 \[ = \dfrac{\pi }{6}\]
So, the correct answer is “Option D”.

Note: Be careful in the calculation part as we can make mistakes in signs. Remember the formula \[{\cos ^{ - 1}}x \pm {\cos ^{ - 1}}y = {\cos ^{ - 1}}\left( {xy \pm \sqrt {1 - {x^2}} \sqrt {1 - {y^2}} } \right)\] so that we can solve these kinds of problems. In the calculation part all we used is basic multiplications, \[\sqrt a \times \sqrt b = \sqrt {ab} \] and \[a\sqrt b = \sqrt {{a^2}b} \] .