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$a \sin x = b \cos x = \frac{2 c \tan x}{1 - \tan^{2} x}$ and ${(a^{2} - b^{2})}^{2} = k c^{2} (a^{2} + b^{2}) \Rightarrow k =$

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Correct option is C

Take asinx=bcosx=ctan2x=lsay⇒sinx=la, cosx=lb, tan2x=lc Now 1=sin2x+cos2x=la2+lb2=l2a2+b2a2b2 =c2tan22xa2+b2a2b2 =c22tanx1-tan2x2a2+b2a2b2 =c22ba1-ba22a2+b2a2b2 ∵tanx=ba ⇒a2-b22=4c2a2+b2 ⇒k=4

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asinx=bcosx=2ctanx1−tan2xand a2−b22=kc2a2+b2⇒k=

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$a \sin x = b \cos x = \frac{2 c \tan x}{1 - \tan^{2} x}$ and ${(a^{2} - b^{2})}^{2} = k c^{2} (a^{2} + b^{2}) \Rightarrow k =$