If tan x = frac{2b}{a - c} (a ne c),y = a cos | vedclass

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(B) Given expressions are $y = a \cos^2 x + 2b \sin x \cos x + c \sin^2 x$ and $z = a \sin^2 x - 2b \sin x \cos x + c \cos^2 x$.Adding $y$ and $z$:$y

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$y + z = a + c$

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(B) Given expressions are $y = a \cos^2 x + 2b \sin x \cos x + c \sin^2 x$ and $z = a \sin^2 x - 2b \sin x \cos x + c \cos^2 x$.Adding $y$ and $z$:$y + z = a(\cos^2 x + \sin^2 x) + c(\sin^2 x + \cos^2 x) = a(1) + c(1) = a + c$.Thus,$y + z = a + c$ is the correct relation.Subtracting $z$ from $y$:$y - z = a(\cos^2 x - \sin^2 x) + 4b \sin x \cos x - c(\cos^2 x - \sin^2 x) = (a - c) \cos 2x + 2b \sin 2x$.Using $	an x = \frac{2b}{a - c}$,we substitute $\cos 2x = \frac{1 - 	an^2 x}{1 + 	an^2 x}$ and $\sin 2x = \frac{2 	an x}{1 + 	an^2 x}$.$y - z = (a - c) \left( \frac{1 - 	an^2 x}{1 + 	an^2 x} ight) + 2b \left( \frac{2 	an x}{1 + 	an^2 x} ight) = \frac{(a - c)(1 - 	an^2 x) + 4b 	an x}{1 + 	an^2 x}$.Substituting $	an x = \frac{2b}{a - c}$:$y - z = \frac{(a - c)(1 - \frac{4b^2}{(a - c)^2}) + 4b(\frac{2b}{a - c})}{1 + \frac{4b^2}{(a - c)^2}} = \frac{(a - c) - \frac{4b^2}{a - c} + \frac{8b^2}{a - c}}{\frac{(a - c)^2 + 4b^2}{(a - c)^2}} = \frac{(a - c) + \frac{4b^2}{a - c}}{\frac{(a - c)^2 + 4b^2}{(a - c)^2}} = \frac{\frac{(a - c)^2 + 4b^2}{a - c}}{\frac{(a - c)^2 + 4b^2}{(a - c)^2}} = a - c$.

If $\tan x = \frac{2b}{a - c}$ $(a \ne c)$,$y = a \cos^2 x + 2b \sin x \cos x + c \sin^2 x$ and $z = a \sin^2 x - 2b \sin x \cos x + c \cos^2 x$,then:

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