Differentiate yourself in trigonometry

$\begin{aligned} y & = \cos x \ln\left(\frac {\sin x+1}{\cos x}\right) \\ & = \cos x \ln (\tan x + \sec x) \\ y' & = - \sin x \ln (\tan x + \sec x) + \cos x \cdot \frac {\sec^2 x + \sec x \tan x}{\tan x + \sec x} \\ & = - y \tan x + \cos x \cdot \sec x \\ \implies y' & = 1 - y \tan x \\ \implies y'' & = - y' \tan x - y \sec^2 x \\ & = - (1-y\tan x) \tan x - y - y \tan^2 x \\ & = - \tan x + y \tan^2 x - y - y\tan^2 x \\ y + y'' & = - \tan x \end{aligned}$

Therefore, $a = \boxed{-1}$ .

• $y=cos(x) \times ln(tan(x) + sec(x))$
• $y'=-sin(x) \times ln(tan(x) + sec(x))$ + $cos(x)$ $\times \frac{sec^{2}(x)+tan(x)sec(x)}{sec(x)+tan(x)}$
• $y'=-sin(x) \times ln(tan(x) + sec(x)) + 1$
• $y''=-cos(x) \times ln(tan(x) + sec(x))-sin(x) \times sec(x)$
• $y' + y'' = cos(x) \times ln(tan(x) + sec(x)) - cos(x) \times ln(tan(x) + sec(x))-sin(x) \times sec(x)$
• = $-tan(x)$
• $a = \boxed{-1}$