Practicing Success
If $sin \theta = \frac{12}{13}$, then $\frac{sin^2 \theta-cos^2\theta}{2cos\theta sin \theta}× cot^2 \theta=?$ |
$\frac{595}{1152}$ $\frac{119}{864}$ $\frac{119}{144}$ $\frac{595}{3456}$ |
$\frac{595}{3456}$ |
sin θ = \(\frac{12}{13}\) { sin θ = \(\frac{P}{H}\) } By using pythagoras theorem, P² + B² = H² 12² + B² = 13² B = 5 Now, \(\frac{sin² θ - co²s θ}{2cosθsinθ}\) × cot² θ = \(\frac{(12/13)² - (5/13)²}{2×5/13×12/13 }\) × ( 5/12)² = \(\frac{144/169 - 25/169}{120/169 }\) × 25/144 = \(\frac{119}{120 }\) × 25/144 = \(\frac{595}{3456 }\) sin2θ−cos2θ2cosθsinθ×cot2θ=? |