Question

The value of \[{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) - {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)\] is
A.\[{45^ \circ }\]
B.\[{90^ \circ }\]
C.\[{15^ \circ }\]
D.\[{30^ \circ }\]

Answer 1

Hint: In the question related to the inverse trigonometric ratios we solve it by using trigonometric ratios values by converting them to required angles of specific values like \[\left( {\sin {{30}^ \circ } = \dfrac{1}{2}} \right)\] . Same method we have to apply here for the given values . You have to remember the values of \[\sin \] for questions related to inverse trigonometry and identities too .

Complete step-by-step answer:
Given : \[{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) - {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)\] .
Now we know that \[\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}\] and \[\sin {30^ \circ } = \dfrac{1}{2}\] , using these values we get ,
\[ = {\sin ^{ - 1}}\left( {\sin {{60}^ \circ }} \right) - {\sin ^{ - 1}}\left( {\sin {{30}^ \circ }} \right)\]
On simplifying we get ,
\[ = {60^ \circ } - {30^ \circ }\] , on solving we get
\[ = {30^ \circ }\]
Therefore , option ( D ) is the correct answer for the given question .
So, the correct answer is “Option D”.

Note: Alternate Method :
Given : \[{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) - {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)\]
We can use the identity of \[{\sin ^{ - 1}}x - {\sin ^{ - 1}}y = {\sin ^{ - 1}}\left[ {x\sqrt {1 - {y^2}} - y\sqrt {1 - {x^2}} } \right]\]
On applying the above identity we get ,
\[ = {\sin ^{ - 1}}\left[ {\dfrac{{\sqrt 3 }}{2}\sqrt {1 - {{\left( {\dfrac{1}{2}} \right)}^2}} - \dfrac{1}{2}\sqrt {1 - {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}} } \right]\] , on simplifying we get
\[ = {\sin ^{ - 1}}\left[ {\dfrac{{\sqrt 3 }}{2}\sqrt {\dfrac{{4 - 1}}{4}} - \dfrac{1}{2}\sqrt {\dfrac{{4 - 3}}{4}} } \right]\]
On further solving we get
\[ = {\sin ^{ - 1}}\left[ {\dfrac{{\sqrt 3 }}{2} \times \dfrac{{\sqrt 3 }}{2} - \dfrac{1}{2} \times \dfrac{1}{2}} \right]\]
\[ = {\sin ^{ - 1}}\left[ {\dfrac{3}{4} - \dfrac{1}{4}} \right]\]
On simplifying we get ,
\[ = {\sin ^{ - 1}}\left[ {\dfrac{1}{2}} \right]\] , we know that \[\sin {30^ \circ } = \dfrac{1}{2}\] , therefore
\[ = {\sin ^{ - 1}}\left[ {\sin {{30}^ \circ }} \right]\]
On simplifying we get the final answer as
\[ = {30^ \circ }\] .
Hence proved .