If sin θ$=\frac{5}{13}$, then what is the value of $\frac{cos^2θ-sin^2θ}{2cosθ.sinθ}$?

119/120

We are given,

sin θ = \(\frac{5}{13}\)

{ we know, sinA = \(\frac{P}{H}\) }

So, P = 5 and H = 13

By using pythagoras theorem,

P² + B² = H²

5² + B² = 13²

B² = 169 - 25 = 144

B = 12

Now,

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

= \(\frac{B²/H² - P² /H² }{2.B/H.P/H}\)

= \(\frac{144- 25 }{2 × 12 × 5 }\)

= \(\frac{119 }{120 }\)