If atan theta = b, then acos 2theta + bsin 2t | vedclass

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Question

(a) Given that $	an 	heta = \frac{b}{a}$.

Now, $a\cos 2	heta + b\sin 2	heta $

$= a\left( {\frac{{1 - {{	an }^2}	heta }}{{1 + {{	an }^2}

Vedclass · Accepted Answer

$a$

Vedclass · Answer

(a) Given that $	an 	heta = \frac{b}{a}$.

Now, $a\cos 2	heta + b\sin 2	heta $

$= a\left( {\frac{{1 - {{	an }^2}	heta }}{{1 + {{	an }^2}	heta }}} ight) + b\left( {\frac{{2	an 	heta }}{{1 + {{	an }^2}	heta }}} ight)$

Putting $	an 	heta = \frac{b}{a}$, we get

$ = a\left( {\frac{{1 - \frac{{{b^2}}}{{{a^2}}}}}{{1 + \frac{{{b^2}}}{{{a^2}}}}}} ight) + b\left( {\frac{{2\frac{b}{a}}}{{1 + \frac{{{b^2}}}{{{a^2}}}}}} ight) $

$= a\left( {\frac{{{a^2} - {b^2}}}{{{a^2} + {b^2}}}} ight) + b\left( {\frac{{2ba}}{{{a^2} + {b^2}}}} ight)$

$ = \frac{1}{{({a^2} + {b^2})}}\{ {a^3} - a{b^2} + 2a{b^2}\} $

$= \frac{{a({a^2} + {b^2})}}{{{a^2} + {b^2}}} = a$.

If $a\tan \theta = b$, then $a\cos 2\theta + b\sin 2\theta = $

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