Question : If $\tan \theta=\frac{4}{5}$, then $\sec \theta$ is:
Option 1: $\frac{\sqrt{41}}{5}$
Option 2: $\frac{\sqrt{41}}{4}$
Option 3: $\frac{5}{4}$
Option 4: $\frac{\sqrt{41}}{20}$
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Correct Answer: $\frac{\sqrt{41}}{5}$
Solution : Given: $\tan \theta=\frac{4}{5}$ We know, $\sec^2 \theta-\tan^2 \theta=1$ $⇒\sec^2 \theta-\frac{16}{25}=1$ $⇒\sec^2 \theta=\frac{41}{25}$ $\therefore\sec \theta=\frac{\sqrt{41}}{5}$ Hence, the correct answer is $\frac{\sqrt{41}}{5}$.
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Question : If $\sqrt{2} \sec ^2 \theta-4 \sec \theta+2 \sqrt{2}=0$, then what is the value $\sin ^2 \theta+\tan ^2 \theta$?
Option 1: $\frac{1}{2}$
Option 2: $\frac{2}{3}$
Option 3: $\frac{5}{2}$
Option 4: $\frac{3}{2}$
Question : If $(r\cos \theta -\sqrt{3})^{2}+(r\sin \theta -1)^{2}=0$, then the value of $\frac{r\tan \theta +\sec \theta}{r\sec \theta +\tan\theta}$ is equal to:
Option 1: $\frac{4}{5}$
Option 2:
$\frac{5}{4}$
Option 3:
$\frac{\sqrt{3}}{4}$
Option 4:
$\frac{\sqrt{5}}{4}$
Question : If $\sec \theta+\tan \theta=\frac{1}{\sqrt{3}}$, then the positive value of $\cot \theta+\cos \theta$ is:
Option 1: $\frac{3 \sqrt{3}}{2}$
Option 2: $\frac{\sqrt{3}}{2}$
Option 3: $\frac{2}{3 \sqrt{3}}$
Option 4: $\frac{2}{\sqrt{3}}$
Question : If $\tan\theta+\sec\theta=3$, $\theta$ being acute, the value of $5\sin\theta$ is:
Option 1: $\frac{5}{2}$
Option 2: $\frac{\sqrt{3}}{5}$
Option 3: $\frac{5}{\sqrt{3}}$
Option 4: $4$
Question : If $\sqrt{3} \tan ^2 \theta-4 \tan \theta+\sqrt{3}=0$, then what is the value of $\tan ^2 \theta+\cot ^2 \theta$?
Option 1: $\frac{4}{3}$
Option 2: $\frac{10}{3}$
Option 3: $3$
Option 4: $\frac{6}{5}$
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