Question

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

• A. $5$
• B. $4$
• C. $-5$
• D. $-4$

Answer 1

Answer: (A), (C)

$-5$

For $\sin x=\frac{12}{13}$
From above triangle
Case $1:$ when $x\in [0,\frac{\pi }{2}]$
$\sec x=\frac{13}{5},\tan x=\frac{12}{5}\Rightarrow \tan x+\sec x=5$

Case $2:$ when $x\in [\frac{\pi }{2},\pi ]$
in this quadrant $\sec x$ and $\tan x$ both are negative
$\therefore \sec x=-\frac{13}{5},\tan x=-\frac{12}{5}\Rightarrow \tan x+\sec x=-5$

Alternate:
As we know
${\cos }^{2}x+{\sin }^{2}x=1\Rightarrow \cos x=\pm \sqrt{1-{\sin }^{2}x}$
Now
Case $1:$ When $x$ lies in $1^{\text{st}}$ quadrant
$\therefore \cos x=\sqrt{1-{\sin }^{2}x}$
and $\sec x+\tan x=\frac{1}{\cos x}+\frac{\sin x}{\cos x}$
$=\frac{1+\sin x}{\cos x}=\frac{1+\sin x}{\sqrt{1-{\sin }^{2}x}}=\frac{1+\frac{12}{13}}{\sqrt{1-{\left(\frac{12}{13}\right)}^2}}=\frac{\frac{25}{13}}{\sqrt{\frac{25}{169}}}=5$

Case $2:$ When $x$ lies in $2^{\text{nd}}$ quadrant
$\therefore \cos x=-\sqrt{1-{\sin }^{2}x}$
and $\sec x+\tan x=\frac{1}{\cos x}+\frac{\sin x}{\cos x}$
$=\frac{1+\sin x}{\cos x}=\frac{1+\sin x}{-\sqrt{1-{\sin }^{2}x}}=\frac{1+\frac{12}{13}}{-\sqrt{1-{\left(\frac{12}{13}\right)}^2}}=\frac{\frac{25}{13}}{-\sqrt{\frac{25}{169}}}=-5$