Question

The number of solutions of the equation 5 sec$\theta$ - 13 = 12 tan$\theta$ in [0 , 2$\pi$] is

• A. 2
• B. 1
• C. 4
• D. 0

Answer 1

The correct option is A

2

5 sec$\theta$ - 13 = 12 tan$\theta$

or, 13 cos$\theta$ + 12 sin $\theta$ = 5

or, $\frac{13}{\sqrt{{13}^2+{12}^2}}$cos$\theta$ + $\frac{12}{\sqrt{{13}^2+{12}^2}}$sin$\theta$ = $\frac{5}{\sqrt{{13}^2+{12}^2}}$

or, cos($\theta$ - $\alpha$) = $\frac{5}{\sqrt{313}}$, where cos$\alpha$ = $\frac{13}{\sqrt{313}}$

$\therefore \theta$ = 2n$\pi$ ± $co{s}^{-1}$ $\frac{5}{\sqrt{313}}$ + $\alpha$

= 2n$\pi$ ± $co{s}^{-1}$ $\frac{5}{\sqrt{313}}$ + $co{s}^{-1}$ $\frac{13}{\sqrt{313}}$

As $co{s}^{-1}$ $\frac{5}{\sqrt{313}}$ > $co{s}^{-1}$ $\frac{13}{\sqrt{313}}$, then

$\theta \in [0,2\pi ]$ , when n = 0 (One value, taking positive sign) and when n = 1 (One value, taking negative sign.)