If $sin \theta = \frac{12}{13}$, then $\frac{sin^2 \theta-cos^2\theta}{2cos\theta sin \theta}× cot^2 \theta=?$
Answer & explanation
Correct answer: option 4
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=?