CUET Preparation Today
If $cosx =\frac{p}{q}$ and 0° < x < 90°, then the value of tanx is : |
$\frac{\sqrt{q^2-p^2}}{q}$ $\frac{q}{\sqrt{q^2-p^2}}$ $\frac{p}{\sqrt{p^2-q^2}}$ $\frac{\sqrt{q^2-p^2}}{p}$ |
$\frac{\sqrt{q^2-p^2}}{p}$ |
cosx = \(\frac{p }{q}\) { cosx = \(\frac{B }{H}\) } B² + P² = H² p² + P² = q² P = \(\sqrt {q² - p² }\) Now, tanx = \(\frac{P }{B}\) = \(\frac{\sqrt {q² - p² }\ }{p}\) |
