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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