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

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(B) Given the equation: $a \cos 2	heta + b \sin 2	heta = c$. Using the half-angle identities $\cos 2	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	h

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

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(B) Given the equation: $a \cos 2	heta + b \sin 2	heta = c$. Using the half-angle identities $\cos 2	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$ and $\sin 2	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta}$,we substitute these into the equation: $a \left( \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} ight) + b \left( \frac{2 	an 	heta}{1 + 	an^2 	heta} ight) = c$. Multiplying both sides by $(1 + 	an^2 	heta)$: $a(1 - 	an^2 	heta) + 2b 	an 	heta = c(1 + 	an^2 	heta)$. Rearranging the terms to form a quadratic equation in $	an 	heta$: $a - a 	an^2 	heta + 2b 	an 	heta = c + c 	an^2 	heta$. $(a + c) 	an^2 	heta - 2b 	an 	heta + (c - a) = 0$. Since $\alpha$ and $\beta$ are the solutions for $	heta$,$	an \alpha$ and $	an \beta$ are the roots of this quadratic equation. Using the sum of roots formula for a quadratic equation $Ax^2 + Bx + C = 0$,where the sum of roots is $-B/A$: $	an \alpha + 	an \beta = - \frac{-2b}{a + c} = \frac{2b}{c + a}$.

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

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