Question

Solve \[\sin \left( {2{{\sin }^{ - 1}}\sqrt {\dfrac{{63}}{{65}}} } \right) = \]
A.\[\dfrac{{2\sqrt {126} }}{{65}}\]
B.\[\dfrac{{4\sqrt {126} }}{{65}}\]
C.\[\dfrac{{8\sqrt {63} }}{{65}}\]
D.\[\dfrac{{63}}{{65}}\]

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 \[\left( {\sin {{30}^ \circ } = \dfrac{1}{2}} \right)\] , but when values of ratios \[\left( {\dfrac{{63}}{{65}}} \right)\] are inconvertible we have use identity according to the questions

Complete step-by-step answer:
Given : \[\sin \left( {2{{\sin }^{ - 1}}\sqrt {\dfrac{{63}}{{65}}} } \right)\] , on rearranging the expression we get ,
\[ = \sin \left( {{{\sin }^{ - 1}}\sqrt {\dfrac{{63}}{{65}}} + {{\sin }^{ - 1}}\sqrt {\dfrac{{63}}{{65}}} } \right)\]
Now , using the identity for \[{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = {\sin ^{ - 1}}\left[ {x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} } \right]\] we get ,
 \[ = \sin \left[ {{{\sin }^{ - 1}}\left( {\sqrt {\dfrac{{63}}{{65}}} \sqrt {1 - {{\left( {\sqrt {\dfrac{{63}}{{65}}} } \right)}^2}} + \sqrt {\dfrac{{63}}{{65}}} \sqrt {1 - {{\left( {\sqrt {\dfrac{{63}}{{65}}} } \right)}^2}} } \right)} \right]\] ,
Remember that the angles are already present under root . So , on solving we get ,
\[ = \sin \left[ {{{\sin }^{ - 1}}\left( {\sqrt {\dfrac{{63}}{{65}}} \sqrt {\dfrac{{65 - 63}}{{65}}} + \sqrt {\dfrac{{63}}{{65}}} \sqrt {\dfrac{{65 - 63}}{{65}}} } \right)} \right]\] , on solving further we get ,
\[ = \sin \left[ {{{\sin }^{ - 1}}\left( {\sqrt {\dfrac{{63}}{{65}}} \sqrt {\dfrac{2}{{65}}} + \sqrt {\dfrac{{63}}{{65}}} \sqrt {\dfrac{2}{{65}}} } \right)} \right]\]
On simplifying we get ,
\[ = \sin \left[ {{{\sin }^{ - 1}}\left( {\dfrac{{\sqrt {126} }}{{65}} + \dfrac{{\sqrt {126} }}{{65}}} \right)} \right]\] , on adding we get
\[ = \sin \left[ {{{\sin }^{ - 1}}\left( {\dfrac{{2\sqrt {126} }}{{65}}} \right)} \right]\]
On simplifying we get ,
\[ = \dfrac{{2\sqrt {126} }}{{65}}\]
Therefore , the option ( A ) is the correct answer for the given option .
So, the correct answer is “Option A”.

Note: Given : \[\sin \left( {2{{\sin }^{ - 1}}\sqrt {\dfrac{{63}}{{65}}} } \right)\]
Now using the identity for \[2{\sin ^{ - 1}} = {\sin ^{ - 1}}2x\sqrt {1 - {x^2}} \] we get
\[ = \sin \left( {{{\sin }^{ - 1}}2\sqrt {\dfrac{{63}}{{65}}} \sqrt {1 - {{\left( {\sqrt {\dfrac{{63}}{{65}}} } \right)}^2}} } \right)\] , on solving we get
\[ = \sin \left( {{{\sin }^{ - 1}}2\sqrt {\dfrac{{63}}{{65}}} \sqrt {\dfrac{{65 - 63}}{{65}}} } \right)\]
On simplifying we get ,
\[ = \sin \left( {{{\sin }^{ - 1}}2\sqrt {\dfrac{{63}}{{65}}} \sqrt {\dfrac{2}{{65}}} } \right)\] , on solving we get
\[ = \sin \left[ {{{\sin }^{ - 1}}\left( {\dfrac{{2\sqrt {126} }}{{65}}} \right)} \right]\]
On rationalizing we get ,
\[ = \dfrac{{2\sqrt {126} }}{{65}}\] .