Question

If $\sin x=\frac{12}{13}$ and $x\in [0,\pi ]$, then the possible value(s) of $\sec x+\tan x$ is/are

• A. $5$
• B. $4$
• C. $-5$
• D. $-4$

Answer 1

Answer: (A), (C)

$-5$

For $\sin x=\frac{12}{13}$
Base $=\sqrt{{13}^2-{12}^2}=5$

Case $1:$ When $x\in [0,\frac{\pi }{2}]$
In first quadrant, both $\sec x$ and $\tan x$ are positive.
$\therefore \sec x=\frac{13}{5},\tan x=\frac{12}{5}\Rightarrow \tan x+\sec x=5$

Case $2:$ When $x\in (\frac{\pi }{2},\pi ]$
In second quadrant, both $\sec x$ and $\tan x$ are negative.
$\therefore \sec x=-\frac{13}{5},\tan x=-\frac{12}{5}\Rightarrow \tan x+\sec x=-5$


Alternate solution:
Given $\sin x=\frac{12}{13}$
$\tan x+\sec x=\frac{\sin x}{\cos x}+\frac{1}{\cos x}=\frac{1+\sin x}{\cos x}=\frac{25}{13\cos x}=\frac{25}{\pm 13\sqrt{1-{\sin }^{2}x}}=\frac{25}{\pm 13\times \frac{5}{13}}=\pm 5$