If $cosec θ=\frac{(x^2+y^2)}{(x^2-y^2)}$, then what will be the value of tan θ ?
Answer & explanation
Correct answer: option 3
cosec θ = \(\frac{x² + y² }{x² - y²}\)
{ we know , cosec θ = \(\frac{H}{P}\) }
By using pythagoras theorem,
P² + B² = H²
(x² - y²)² + B² = (x² + y²)²
On solving it,
B = 2xy
Now,
tan θ = \(\frac{P}{B}\)
= \(\frac{x² - y²}{2xy}\)