If $sin \theta = \frac{2\sqrt{ab}}{a+b}, a > b > 0, $ then the value of $\frac{cos\theta + 1}{cos\theta - 1}$ will be :

$-\frac{a}{b}$

sin θ = \(\frac{2√ab}{a + b }\)

{ sin θ = \(\frac{P}{H }\) }

By using pythagoras theorem,

P² + B² = H²

4ab + B² = (a+b)²

B² = a² + b² + 2ab - 4ab

B = a - b

Now,

\(\frac{cosθ + 1 }{cosθ - 1 }\)

= \(\frac{(a-b)/(a+b)+ 1 }{(a-b)/(a+b) - 1 }\)

= \(\frac{a-b+ a+b }{a - b - a -b }\)

= - \(\frac{a}{b }\)