If $cosx =\frac{p}{q}$ and 0° < x

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Target Exam

CUET

Subject

General Test

Chapter

Quantitative Reasoning

Topic

Trigonometry

Question:

If $cosx =\frac{p}{q}$ and 0° < x < 90°, then the value of tanx is :

Options:

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

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

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

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

Correct Answer:

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

Explanation:

cosx = $\frac{p }{q}$

{ cosx = $\frac{B }{H}$ }

B² + P² = H²

p² + P² = q²

P = $\sqrt {q² - p² }$

Now,

tanx

= $\frac{P }{B}$

= $\frac{\sqrt {q² - p² }\ }{p}$