If X and Y are acute …
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Naveen Sharma 8 years, 5 months ago
Ans. Given : {tex}sin X = {1\over \sqrt 5}{/tex}, {tex}sin Y = {1\over \sqrt {10}}{/tex}
X = {tex}sin^{-1}({1\over \sqrt 5}){/tex}
Y = {tex}sin^{-1}({1\over \sqrt {10}}){/tex}
=> X+Y = {tex}sin^{-1}({1\over \sqrt 5}) + sin^{-1}({1\over \sqrt {10}}){/tex}
{tex}[using \ \ sin^{-1} a + sin^{-1} b = sin^{-1}(a\sqrt{1-b^2}+b\sqrt{1-a^2})]{/tex}
{tex}=> X+Y = sin^{-1}\left ( {1\over \sqrt 5} \sqrt {1- {1\over 10}} + {1\over \sqrt {10}} \sqrt {1- {1\over 5}}\right ){/tex}
{tex}=> X+Y = sin^{-1}\left ( {3\over \sqrt {50}} + {2\over \sqrt {50}} \right ){/tex}
{tex}=> X+Y = sin^{-1}\left ( {5\over 5\sqrt 2} \right ){/tex}
{tex}=> X+Y = sin^{-1}\left ( {1\over \sqrt 2} \right ){/tex}
{tex}=> X+Y = sin^{-1}\left ( sin {\pi \over 4} \right ){/tex}
{tex}=> X+Y = {\pi \over 4} {/tex}
Hence Proved
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