If alpha , beta are different values of x sat | vedclass

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Question

(c) $a\cos x + b\sin x = c$

$ \Rightarrow $ $a\,\left( {\frac{{1 - {{	an }^2}\,(x/2)}}{{1 + {{	an }^2}(x/2)}}} ight) + \frac{{2b	an (x/2)}}{{1

Vedclass · Accepted Answer

$\frac{b}{a}$

Vedclass · Answer

(c) $a\cos x + b\sin x = c$

$ \Rightarrow $ $a\,\left( {\frac{{1 - {{	an }^2}\,(x/2)}}{{1 + {{	an }^2}(x/2)}}} ight) + \frac{{2b	an (x/2)}}{{1 + {{	an }^2}(x/2)}} = c$

$ \Rightarrow $ $(a + c){	an ^2}\frac{x}{2} - 2b	an \frac{x}{2} + (c - a) = 0$

This equation has roots $	an \frac{\alpha }{2}$ and $	an \frac{\beta }{2}$.

$	herefore $ $	an \frac{\alpha }{2} + 	an \frac{\beta }{2} = \frac{{2b}}{{a + c}}$ and

$	an \frac{\alpha }{2}	an \frac{\beta }{2} = \frac{{c - a}}{{a + c}}$

Now $	an \left( {\frac{\alpha }{2} + \frac{\beta }{2}} ight) = \frac{{	an \frac{\alpha }{2} + 	an \frac{\beta }{2}}}{{1 - 	an \frac{\alpha }{2}	an \frac{\beta }{2}}}$

$= \frac{{\frac{{2b}}{{a + c}}}}{{1 - \frac{{c - a}}{{a + c}}}} = \frac{b}{a}$.

If $\alpha ,$ $\beta$ are different values of $x$ satisfying $a\cos x + b\sin x = c,$ then $\tan {\rm{ }}\left( {\frac{{\alpha + \beta }}{2}} \right) = $

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