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: $\frac{3}{4}$
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Correct Answer: $-\frac{5}{4}$
Solution : $\sin A=-\frac{3}{5}$ So, $\cos A = \sqrt{1-\sin^2 A}$ ⇒ $\cos A= \sqrt{1-(\frac{3}{5})^2}$ ⇒ $\cos A = \sqrt\frac{16}{25}$ $\therefore\cos A= \pm \frac{4}{5}$ Since it is 3rd quadrant, $\cos A = -\frac{4}{5}$ $\therefore\sec A = -\frac{5}{4}$ Hence, the correct answer is $-\frac{5}{4}$.
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Question : If $\sec A=\frac{5}{4}$, then the value of $\frac{\tan A}{1+\tan ^2 A}-\frac{\sin A}{\sec A}$ is:
Option 1: 2
Option 2: 1
Option 3: 0
Option 4: 3
Question : If $\frac{\sin ^2 \theta}{\cos ^2 \theta-3 \cos \theta+2}=1, \theta$ lies in the first quadrant, then the value of $\frac{\tan ^2 \frac{\theta}{2}+\sin ^2 \frac{\theta}{2}}{\tan \theta+\sin \theta}$ is:
Option 1: $\frac{2 \sqrt{3}}{27}$
Option 2: $\frac{5 \sqrt{3}}{27}$
Option 3: $\frac{2 \sqrt{3}}{9}$
Option 4: $\frac{7 \sqrt{3}}{54}$
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: $\frac{37}{19}$
Question : If $0°<A<90°$, the value of $\frac{\tan A\ -\ \sec A\ -\ 1}{\tan A\ +\ \sec A\ +\ 1}$ is:
Option 1: $\frac{\sin A-1}{\cos A}$
Option 2: $\frac{1-\sin A}{\cos A}$
Option 3: $\frac{1-\cos A}{\sin A}$
Option 4: $\frac{\sin A+1}{\cos A}$
Question : If $A+B=90^{\circ}$ and $\sin A=\frac{3}{5}$, then the value of $\tan B$ is:
Option 1: $\frac{4}{3}$
Option 2: $\frac{5}{4}$
Option 3: $\frac{3}{4}$
Option 4: $\frac{5}{3}$
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