# Trigonometric Function and Solving Trigonometric Equation

Let $\displaystyle f(x)=\frac{{\cos x}}{{1+\sin x}}+\frac{{1+\sin x}}{{\cos x}}.$

(a) Prove that $\displaystyle f(x)=2\sec x.$

(b) Hence solve the equation $\displaystyle f(x)=4$ where 0 < x < 2π.

Solution

(a)  $\displaystyle f(x)=\frac{{\cos x}}{{1+\sin x}}+\frac{{1+\sin x}}{{\cos x}}$

$\displaystyle \ \ \ \ \ \ \ =\frac{{{{{\cos }}^{2}}x+{{{(1+\sin x)}}^{2}}}}{{(1+\sin x)\cos x}}$

$\displaystyle \ \ \ \ \ \ \ =\frac{{{{{\cos }}^{2}}x+1+2\sin x+\sin^{2}{{x}}}}{{(1+\sin x)\cos x}}$

$\displaystyle \ \ \ \ \ \ \ =\frac{{\sin {{x}^{2}}+{{{\cos }}^{2}}x+1+2\sin x}}{{(1+\sin x)\cos x}}$

$\displaystyle \ \ \ \ \ \ \ =\frac{{1+1+2\sin x}}{{(1+\sin x)\cos x}}$

$\displaystyle \ \ \ \ \ \ \ =\frac{{2(1+\sin x)}}{{(1+\sin x)\cos x}}$

$\displaystyle \ \ \ \ \ \ \ =\frac{2}{{\cos x}}$

$\displaystyle \ \ \ \ \ \ \ =2\sec x$

(b)    $\displaystyle \ f(x)=4$

$\displaystyle \therefore 2\sec x=4$

$\displaystyle \ \ \ \sec x=2$

$\displaystyle \ \ \ x=\frac{\pi }{3}$ or $\displaystyle \ \ \ x=\frac{5\pi }{3}$