Question

if x sin θ = y cos θ = 2z tan θ / 1-tan² θ, then 4z²(x²+y²) is equal to...........

(a) (x²+y²)³
(b) (x²-y²)³
(c) (x²-y²)²
(d) (x²+y²)²

Answer 1

Let k = x cos θ = y sinθ = z *(2sinθ cosθ)/(cOS^2 θ - sin^2 θ) z = k*(cos/sin - sin/cos)/2. ..
z = k(y/x - x/y)/2 = k(y^2-x^2)/2xy....
LHS = k^2 (y^2-x^2)^2 * (x^2+y^2)/(x^2y^2).....
substitute for x and y in terms of k and theta.
x = k /cosθ... y= k/sinθ.
Then (x^2+y^2)/x^2y^2 = 1/k^2.
LHS = (x^2 - y^2)^2