Question:
If $\sin x=\frac{12}{13}$ and $x$ lies in the second quadrant, find the value of $\sec x+\tan x$.
Solution:
We have:
$\sin x=\frac{12}{13}$ and $x$ lie in the second quadrant.
In the second quadrant, $\sin x$ and $\operatorname{cosec} x$ are positive and all the other four $\mathrm{T}-$ ratios are negative.
$\therefore \cos x=-\sqrt{1-\sin ^{2} x}$
$=-\sqrt{1-\left(\frac{12}{13}\right)^{2}}$
$=\frac{-5}{13}$
$\tan x=\frac{\sin x}{\cos x}$
$=\frac{\frac{12}{13}}{\frac{-5}{13}}$
$=\frac{-12}{5}$
And, $\sec x=\frac{1}{\cos x}$
$=\frac{1}{\frac{-5}{13}}$
$=\frac{-13}{5}$
$\therefore \sec x+\tan x=\frac{-13}{5}+\frac{-12}{5}$
$=-5$
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