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Posted by Simran Prajapati 4 years, 2 months ago
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Preeti Dabral 4 years, 2 months ago
TO DETERMINE
CALCULATION
It is given that
Now a relation from A to A is a subset of A × A
Where A × A is the Cartesian product of A and A
We we have to determine a Relation R with the below mentioned property
Now there does not exist any ( a, b) in A × A such that a - b = 10
So the Required Relation is Empty
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Preeti Dabral 4 years, 3 months ago
Let {tex}{\cos ^{ - 1}}\frac{{12}}{{13}} = \theta{/tex} so that {tex}\cos \theta = \frac{{12}}{{13}}{/tex}
{tex}\therefore \sin \theta = \sqrt {1 - {{\cos }^2}\theta } = \sqrt {1 - \frac{{144}}{{169}}}{/tex}{tex}= \sqrt {\frac{{25}}{{169}}} = \frac{5}{{13}}{/tex}
Again, Let {tex}{\sin ^{ - 1}}\frac{3}{5} = \phi{/tex} so that {tex}\sin \phi = \frac{3}{5}{/tex}
{tex}\therefore \cos \phi = \sqrt {1 - {{\sin }^2}\phi } = \sqrt {1 - \frac{9}{{25}}} = \sqrt {\frac{{16}}{{25}}} = \frac{4}{5}{/tex}
Since {tex}\sin \left( {\theta + \phi } \right) = \sin \theta \cos \phi + \cos \theta \sin \phi {/tex} {tex} = \frac{5}{{13}} \times \frac{4}{5} + \frac{{12}}{{13}} \times \frac{3}{5}{/tex}
{tex}= \frac{{20 + 36}}{{65}} = \frac{{56}}{{65}}{/tex}
{tex}\Rightarrow \theta + \phi = {\sin ^{ - 1}}\frac{{56}}{{65}}{/tex}
{tex} \Rightarrow {\cos ^{ - 1}}\frac{{12}}{{13}} + {\sin ^{ - 1}}\frac{3}{5} = {\sin ^{ - 1}}\frac{{56}}{{65}}{/tex}
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Siddhant Singh 4 years, 2 months ago
0Thank You