Question

If $\frac{\sin x+\cos x}{{\cos }^{3}x}=a{\tan }^{3}x+b{\tan }^{2}x+c\tan x+d$ then $a+b+c+d=$

• A. $0$
• B. $2$
• C. $4$
• D. $-2$

Answer 1

The correct option is C $4$

$\frac{sinx+cosx}{co{s}^{3}x}=ata{n}^{3}x+bta{n}^{2}x+ctanx+d$

To find; $a+b+c+d$, Put, $x=\frac{\pi }{4}$

We get,

=> $\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{\frac{1}{2\sqrt{2}}}=a+b+c+d$

=> $\sqrt{2}\times 2\sqrt{2}=a+b+c+d$

Hence,

=> $a+b+c+d=4$