If \cos x = \frac{p}{q} and 0^\circ < x < 90^\circ, then the value of \tan x is:

Question 55

If $$\cos x = \frac{p}{q}$$ and $$0^\circ < x < 90^\circ$$, then the value of $$\tan x$$ is:

Solution

Given, $$\cos x = \frac{p}{q}$$

$$\therefore\ $$ $$\tan x=\frac{\sin x}{\cos x}$$

$$=\frac{\sqrt{1-\cos^2x}}{\cos x}$$

$$=\frac{\sqrt{1-\left(\frac{p}{q}\right)^2}}{\frac{p}{q}}$$

$$=\frac{\sqrt{\frac{q^2-p^2}{q^2}}}{\frac{p}{q}}$$

$$=\frac{\frac{\sqrt{q^2-p^2}}{q}}{\frac{p}{q}}$$

$$=\frac{\sqrt{q^2-p^2}}{p}$$

Hence, the correct answer is Option B

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