Question : If $\tan x = –\frac{12}{5}$, where $x$ lies in the second quadrant, what is the value of $\sin x-\cot x$?
Option 1: $\frac{209}{156}$
Option 2: $\frac{169}{156}$
Option 3: $\frac{156}{209}$
Option 4: $\frac{144}{169}$
Correct Answer: $\frac{209}{156}$
Solution :
In the second quadrant, $\sin x$ and $\operatorname{cosec x}$ are positive.
$\tan x = –\frac{12}{5}$
$⇒\cot x=\frac{1}{\tan x}=-\frac{5}{12}$
We know, $\sec^2x -\tan^2x=1$
$⇒\sec^2x=1+(–\frac{12}{5})^2$
$⇒\sec^2x=1+\frac{144}{25}$
$⇒\sec^2x=\frac{169}{25}$
$⇒ \sec x=\frac{13}{5}$
$⇒\cos x = \frac{5}{13}$
$\therefore\sin x = \sqrt{1-\cos^2x}=\sqrt{1-\frac{25}{169}}=\frac{144}{169}=\frac{12}{13}$
$\sin x - \cot x = \frac{12}{13} – (-\frac{5}{12}) = \frac{144 + 65}{156} = \frac{209}{156}$
Hence, the correct answer is $\frac{209}{156}$.
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