If sinθ = $\frac{12}{13}, 0 < θ < 90^o, $ then $\frac{sin^2θ-cos^2θ}{2sinθ.cosθ}×\frac{1}{tan^2θ}$ = ____________.

$\frac{595}{3456}$

sinθ = \(\frac{12}{13}\)

{ we know, sinθ = \(\frac{P}{H}\) }

By using pythagoras theorem,

P² + B² = H²

12² + B² = 13²

B = 5

Now,

\(\frac{ sin²θ - cos²θ }{2sinθ.cosθ}\) × \(\frac{1}{tan²θ}\)

= \(\frac{ P² - B² }{2P.B}\)× \(\frac{B²}{P²}\)

= \(\frac{ 12² - 5² }{2× 12×5 }\)× \(\frac{5²}{12²}\)

= \(\frac{ 144 - 25 }{120 }\)× \(\frac{25}{144}\)

= \(\frac{ 2975 }{17280 }\)

= \(\frac{ 595 }{ 3456 }\)