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Question 55
If $$\cos x = \frac{p}{q}$$ and $$0^\circ < x < 90^\circ$$, then the value of $$\tan x$$ is:
Solution
Given, Â $$\cos x = \frac{p}{q}$$
$$\therefore\ $$ $$\tan x=\frac{\sin x}{\cos x}$$
$$=\frac{\sqrt{1-\cos^2x}}{\cos x}$$
$$=\frac{\sqrt{1-\left(\frac{p}{q}\right)^2}}{\frac{p}{q}}$$
$$=\frac{\sqrt{\frac{q^2-p^2}{q^2}}}{\frac{p}{q}}$$
$$=\frac{\frac{\sqrt{q^2-p^2}}{q}}{\frac{p}{q}}$$
$$=\frac{\sqrt{q^2-p^2}}{p}$$
Hence, the correct answer is Option B
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