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: $\frac{144}{169}$
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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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Question : Using trigonometric formulas, find the value of $(\frac{\sin (x-y)}{\sin (x+y)})(\frac{\tan x+\tan y}{\tan x-\tan y})$
Option 1: –2
Option 2: 2
Option 3: 0
Option 4: 1
Question : If $\sin A=-\frac{3}{5}, A$ lies in III quadrant, the value of $\sec A$ is:
Option 1: $-\frac{4}{5}$
Option 2: $-\frac{5}{4}$
Option 3: $-\frac{3}{4}$
Option 4: $\frac{3}{4}$
Question : If $\tan x = \frac{7}{5}$, the value of $\frac{9 \sin x – \frac{42}{5} \cos x}{15 \sin x + 21 \cos x}$ is:
Option 1: 0
Option 2: 1
Option 3: 0.1
Option 4: 0.5
Question : If $\frac{\sin x-\cos x}{\sin x+\cos x}=\frac{2}{5}$, then the value of $\frac{1+\cot ^2 x}{1-\cot ^2 x}$ is:
Option 1: 2.25
Option 2: 1.45
Option 3: 3.75
Option 4: 5.25
Question : If $\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}=\frac{1}{7}, \theta$ lies in first quadrant, then the value of $\frac{\operatorname{cosec} \theta+\cot ^2 \theta}{\operatorname{cosec} \theta-\cot ^2 \theta}$ is:
Option 1: $\frac{19}{5}$
Option 2: $\frac{22}{3}$
Option 3: $\frac{37}{12}$
Option 4: $\frac{37}{19}$
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