Solveeit Logo
Login

Question

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

The value of sin⁡−1(32)−sin⁡−1(12){\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) - {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) is
A.45∘{45^ \circ }
B.90∘{90^ \circ }
C.15∘{15^ \circ }
D.30∘{30^ \circ }

Explanation

Solution

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 (sin⁡30∘=12)\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⁡\sin for questions related to inverse trigonometry and identities too .

Complete step-by-step answer :
Given : sin⁡−1(32)−sin⁡−1(12){\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) - {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) .
Now we know that sin⁡60∘=32\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2} and sin⁡30∘=12\sin {30^ \circ } = \dfrac{1}{2} , using these values we get ,
=sin⁡−1(sin⁡60∘)−sin⁡−1(sin⁡30∘)= {\sin ^{ - 1}}\left( {\sin {{60}^ \circ }} \right) - {\sin ^{ - 1}}\left( {\sin {{30}^ \circ }} \right)
On simplifying we get ,
=60∘−30∘= {60^ \circ } - {30^ \circ } , on solving we get
=30∘= {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(32)−sin⁡−1(12){\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) - {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)
We can use the identity of sin⁡−1x−sin⁡−1y=sin⁡−1[x1−y2−y1−x2]{\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[321−(12)2−121−(32)2]= {\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[324−14−124−34]= {\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[32×32−12×12]= {\sin ^{ - 1}}\left[ {\dfrac{{\sqrt 3 }}{2} \times \dfrac{{\sqrt 3 }}{2} - \dfrac{1}{2} \times \dfrac{1}{2}} \right]
=sin⁡−1[34−14]= {\sin ^{ - 1}}\left[ {\dfrac{3}{4} - \dfrac{1}{4}} \right]
On simplifying we get ,
=sin⁡−1[12]= {\sin ^{ - 1}}\left[ {\dfrac{1}{2}} \right] , we know that sin⁡30∘=12\sin {30^ \circ } = \dfrac{1}{2} , therefore
=sin⁡−1[sin⁡30∘]= {\sin ^{ - 1}}\left[ {\sin {{30}^ \circ }} \right]
On simplifying we get the final answer as
=30∘= {30^ \circ } .
Hence proved .