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 ,
$= \dfrac{{2\sqrt {126} }}{{65}}$ .