If Sinx+CosxCos3x=a tan3x + b tan2x+c tan x+d then a+b+c+d=

sin x+cos xcos3x=a tan3x+b tan2x+c tan x+dtan x=1 ⇒ x=π412+12122=a+b+c+da+b+c+d=4

If Sinx+CosxCos3x=a tan3x + b tan2x+c tan x+d then a+b+c+d=

sin x+cos xcos3x=a tan3x+b tan2x+c tan x+dtan x=1 ⇒ x=π412+12122=a+b+c+da+b+c+d=4

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If $\frac{Sinx + Cosx}{{Cos}^{3} x} = a \tan^{3} x + b \tan^{2} x + c \tan x + d$ then $a + b + c + d =$

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Detailed Solution

$\begin{array}{rcl} \frac{\sin x + \cos x}{\cos^{3} x} & = & a \tan^{3} x + b \tan^{2} x + c \tan x + d \\ \tan x & = & 1 \Rightarrow x = \frac{π}{4} \\ \frac{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}}{\frac{1}{2 \sqrt{2}}} & = & a + b + c + d \\ a + b + c + d & = & 4 \end{array}$