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Question: The value of \[{\cos ^{ - 1}}\left( {\sqrt {\dfrac{2}{3}} } \right) - {\cos ^{ - 1}}\left( {\dfrac{{...

The value of cos⁡−1(23)−cos⁡−1(6+123){\cos ^{ - 1}}\left( {\sqrt {\dfrac{2}{3}} } \right) - {\cos ^{ - 1}}\left( {\dfrac{{\sqrt 6 + 1}}{{2\sqrt 3 }}} \right) is equal to
A. π3\dfrac{\pi }{3}
B. π4\dfrac{\pi }{4}
C. π2\dfrac{\pi }{2}
D. π6\dfrac{\pi }{6}

Explanation

Solution

Hint : We use the formula cos⁡−1x±cos⁡−1y=cos⁡−1(xy±1−x21−y2){\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=a2+b2+2ab{(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(23)−cos⁡−1(6+123){\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⁡−1x−cos⁡−1y=cos⁡−1(xy−1−x21−y2){\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=23x = \sqrt {\dfrac{2}{3}} and y=6+123y = \dfrac{{\sqrt 6 + 1}}{{2\sqrt 3 }}
Then,
=cos⁡−1((23×6+123)−1−(23)21−(6+123)2)= {\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 (23×6+123)\left( {\sqrt {\dfrac{2}{3}} \times \dfrac{{\sqrt 6 + 1}}{{2\sqrt 3 }}} \right) and 1−(23)21−(6+123)2\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, (23×6+123)\left( {\sqrt {\dfrac{2}{3}} \times \dfrac{{\sqrt 6 + 1}}{{2\sqrt 3 }}} \right)
=23×6+123= \dfrac{{\sqrt 2 }}{{\sqrt 3 }} \times \dfrac{{\sqrt 6 + 1}}{{2\sqrt 3 }}
Using simple multiplication, we get:
=2×(6+1)2×3= \dfrac{{\sqrt 2 \times (\sqrt 6 + 1)}}{{2 \times 3}}
Expanding the brackets in the numerator,
=2×6+26= \dfrac{{\sqrt {2 \times 6} + \sqrt 2 }}{6}
=12+26= \dfrac{{\sqrt {12} + \sqrt 2 }}{6}
We can again 12 has 12=4×312 = 4 \times 3
=4×3+26= \dfrac{{\sqrt {4 \times 3} + \sqrt 2 }}{6}
We know the square root of 4 is 2.
=23+26= \dfrac{{2\sqrt 3 + \sqrt 2 }}{6} ------- (2).
Now, take 1−(23)21−(6+123)2\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:
=1−231−(6+1)2(23)2= \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=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab ,
=3−231−(6)2+12+264×3= \sqrt {\dfrac{{3 - 2}}{3}} \sqrt {1 - \dfrac{{{{\left( {\sqrt 6 } \right)}^2} + {1^2} + 2\sqrt 6 }}{{4 \times 3}}}
=131−(6+1+26)12= \dfrac{1}{{\sqrt 3 }}\sqrt {1 - \dfrac{{(6 + 1 + 2\sqrt 6 )}}{{12}}}
Taking L.C.M. and simplifying,
=1312−7−2612= \dfrac{1}{{\sqrt 3 }}\sqrt {\dfrac{{12 - 7 - 2\sqrt 6 }}{{12}}}
=135−2612= \dfrac{1}{{\sqrt 3 }}\sqrt {\dfrac{{5 - 2\sqrt 6 }}{{12}}}
=(5−26)6= \dfrac{{\sqrt {(5 - 2\sqrt 6 )} }}{6}
We will simplify it further so that square root will get cancel, that is (2−3)2=2+3−226=5−26{\left( {\sqrt 2 - \sqrt 3 } \right)^2} = 2 + 3 - 2\sqrt 2 \sqrt 6 = 5 - 2\sqrt 6
Substituting in above we get,
=(2−3)26= \dfrac{{\sqrt {{{\left( {\sqrt 2 - \sqrt 3 } \right)}^2}} }}{6}
=(2−3)6= \dfrac{{\left( {\sqrt 2 - \sqrt 3 } \right)}}{6} ----- (3).
Now substituting (2) and (3) in the equation (1) we get:
=cos⁡−1(23+26−(2−3)6)= {\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(23+2−2+36)= {\cos ^{ - 1}}\left( {\dfrac{{2\sqrt 3 + \sqrt 2 - \sqrt 2 + \sqrt 3 }}{6}} \right)
Cancelling 2\sqrt 2 we get,
=cos⁡−1(23+36)= {\cos ^{ - 1}}\left( {\dfrac{{2\sqrt 3 + \sqrt 3 }}{6}} \right)
Taking 3\sqrt 3 as common, we get,
=cos⁡−1(3(2+1)6)= {\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 (2 + 1)}}{6}} \right)
=cos⁡−1(336)= {\cos ^{ - 1}}\left( {\dfrac{{3\sqrt 3 }}{6}} \right)
Cancelling, we get:
=cos⁡−1(32)= {\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)
We know that cos⁡−1(32)=π6{\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = \dfrac{\pi }{6} , so we get:
=π6= \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⁡−1x±cos⁡−1y=cos⁡−1(xy±1−x21−y2){\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, a×b=ab\sqrt a \times \sqrt b = \sqrt {ab} and ab=a2ba\sqrt b = \sqrt {{a^2}b} .