Question : If $\cos\theta=\frac{p}{\sqrt{p^{2}+q^{2}}}$, then the value of $\tan\theta$ is:
Option 1: $\frac{q}{\sqrt{p^{2}-q^{2}}}$
Option 2: $\frac{q}{p}$
Option 3: $\frac{p}{p^{2}+q^{2}}$
Option 4: $\frac{q}{\sqrt{p^{2}+q^{2}}}$
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Correct Answer: $\frac{q}{p}$
Solution : Given that $\cos\theta=\frac{p}{\sqrt{p^{2}+q^{2}}}$. ⇒ $\cos^2\theta =\frac{p^2}{p^2+q^2} $ Using, $\sin^2\theta + \cos^2\theta = 1$ ⇒ $\sin^2\theta = 1-\cos^2\theta$ ⇒ $\sin^2\theta = 1-\frac{p^2}{p^2+q^2}$ ⇒ $\sin^2\theta = \frac{q^2}{p^2+q^2}$ ⇒ $\sin\theta=\frac{q}{\sqrt{p^{2}+q^{2}}}$ ⇒ $\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac{q}{\sqrt{p^{2}+q^{2}}}}{\frac{p}{\sqrt{p^{2}+q^{2}}}}=\frac{q}{p}$ Hence, the correct answer is $\frac{q}{p}$.
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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 $\sqrt{3} \tan ^2 \theta-4 \tan \theta+\sqrt{3}=0$, 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: $\frac{6}{5}$
Question : If $\theta$ is an acute angle and $\sin \theta \cos \theta=2 \cos ^3 \theta-\frac{1}{4} \cos \theta$, then the value of $\sin \theta$ is:
Option 1: $\frac{\sqrt{15}-1}{8}$
Option 2: $\frac{\sqrt{15}-1}{4}$
Option 3: $\frac{\sqrt{15}+1}{4}$
Option 4: $\frac{\sqrt{15}-1}{2}$
Question : If $\theta$ is a positive acute angle and $4\cos ^{2}\theta -4\cos \theta +1=0$, then the value of $\tan (\theta -15^{\circ})$is equal to:
Option 1: $0$
Option 2: $1$
Option 3: $\sqrt{3}$
Option 4: $\frac{1}{\sqrt{3}}$
Question : If $\sin\theta+\cos\theta=\sqrt{2}\cos\theta$, then the value of $\cot\theta$ is:
Option 1: $\sqrt{2}+1$
Option 2: $\sqrt{2}-1$
Option 3: $\sqrt{3}-1$
Option 4: $\sqrt{3}+1$
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