STD 11 - 3. trignometrical ratios,functions and identities — JEE कक्षा 11 साइन्स — Question

अब $a\cos 2\theta  + b\sin 2\theta  $

$= a\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right) + b\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$\tan \theta  = \frac{b}{a}$ रखने पर,

$ = a\left( {\frac{{1 - \frac{{{b^2}}}{{{a^2}}}}}{{1 + \frac{{{b^2}}}{{{a^2}}}}}} \right) + b\left( {\frac{{2\frac{b}{a}}}{{1 + \frac{{{b^2}}}{{{a^2}}}}}} \right) $

$= a\left( {\frac{{{a^2} - {b^2}}}{{{a^2} + {b^2}}}} \right) + b\left( {\frac{{2ba}}{{{a^2} + {b^2}}}} \right)$

$ = \frac{1}{{({a^2} + {b^2})}}\{ {a^3} - a{b^2} + 2a{b^2}\} $

$= \frac{{a({a^2} + {b^2})}}{{{a^2} + {b^2}}} = a$.