If a,cos 2theta + b,sin 2theta = c has | vedclass

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(b) $a\cos 2	heta + b\sin 2	heta = c$ 

==> $a\left( {\frac{{1 - {{	an }^2}	heta }}{{1 + {{	an }^2}	heta }}} ight) + b\frac{{2	an \

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$\frac{{2b}}{{c + a}}$

Vedclass · Answer

(b) $a\cos 2	heta + b\sin 2	heta = c$ 

==> $a\left( {\frac{{1 - {{	an }^2}	heta }}{{1 + {{	an }^2}	heta }}} ight) + b\frac{{2	an 	heta }}{{1 + {{	an }^2}	heta }} = c$ 

$ \Rightarrow $ $a - a{	an ^2}	heta + 2b	an 	heta = c + c{	an ^2}	heta $

$ \Rightarrow $$ - (a + c){	an ^2}	heta + 2b\,	an 	heta + (a - c) = 0$

$	herefore 	an \alpha + 	an \beta = - \frac{{2b}}{{ - (c + a)}} = \frac{{2b}}{{c + a}}$ .

If $a\,\cos 2\theta + b\,\sin 2\theta = c$ has $\alpha$ and $\beta$ as its solution, then the value of $\tan \alpha + \tan \beta $ is

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