If tan⁡x=2ba−c, a≠c, y=acos2⁡x+2bsin⁡x⋅cos⁡x+csin2⁡x; z=asin2⁡x−2bsin⁡x⋅cos⁡x+ccos2⁡x, then

y−z=(a−c)cos⁡2x+2b(sin⁡2x)then substitute tan⁡x=2ba−cy-z=a-c1-tan2x1+tan2x+2b2tanx1+tan2x =a-c1-2ba-c21+2ba-c2+2b22ba-c1+2ba-c2 =a-ca-c2-4b2a-c2+4b2+2b4b(a-c)a-c2+4b2 =a-ca-c2-4b2+8b2a-c2+4b2=a-c

If tan⁡x=2ba−c, a≠c, y=acos2⁡x+2bsin⁡x⋅cos⁡x+csin2⁡x; z=asin2⁡x−2bsin⁡x⋅cos⁡x+ccos2⁡x, then

y−z=(a−c)cos⁡2x+2b(sin⁡2x)then substitute tan⁡x=2ba−cy-z=a-c1-tan2x1+tan2x+2b2tanx1+tan2x =a-c1-2ba-c21+2ba-c2+2b22ba-c1+2ba-c2 =a-ca-c2-4b2a-c2+4b2+2b4b(a-c)a-c2+4b2 =a-ca-c2-4b2+8b2a-c2+4b2=a-c

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If $\tan x = \frac{2 b}{a - c}, a \neq c, y = {acos}^{2} x + 2 bsin x \cdot \cos x + {csin}^{2} x; z = {asin}^{2} x - 2 bsin x \cdot \cos x + {ccos}^{2} x$ , then

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$y - z = a - c$

$y - z = (a - c)^{2} + 4 b^{2}$

$y - z = a + c$

$y = z$

answer is C.

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Detailed Solution

$y - z = (a - c) \cos 2 x + 2 b (\sin 2 x)$

then substitute $\tan x = \frac{2 b}{a - c}$

$y - z = (a - c) (\frac{1 - \tan^{2} x}{1 + \tan^{2} x}) + 2 b (\frac{2 \tan x}{1 + \tan^{2} x}) = (a - c) (\frac{1 - {(\frac{2 b}{a - c})}^{2}}{1 + {(\frac{2 b}{a - c})}^{2}}) + 2 b (\frac{2 (\frac{2 b}{a - c})}{1 + {(\frac{2 b}{a - c})}^{2}}) = (a - c) (\frac{{(a - c)}^{2} - 4 b^{2}}{{(a - c)}^{2} + 4 b^{2}}) + 2 b (\frac{4 b (a - c)}{{(a - c)}^{2} + 4 b^{2}}) = (a - c) (\frac{{(a - c)}^{2} - 4 b^{2} + 8 b^{2}}{{(a - c)}^{2} + 4 b^{2}}) = a - c$