If sin x (frac{P – Q}{P + Q}), where 0(^o) (leq) x (leq) 90(^o), find 1 – tan(^2)x
Explanation
Let w be the adjacent side, then w(^2) = (P + q)(^2) = 4pq so that w = 2(sqrt{pq})
Since tan x (frac{p – q}{2sqrt{pq}}),
substituting; 1 – tan(^2)x = 1 – ((frac{p – q}{2sqrt{pq}}))(^2)
= 1 – (frac{P^2 – 2pq + q^2}{4pq})
= (frac{4pq – p^2 + 2pq – q^2}{4pq})
Therefore, 1 – tan(^2)x = (frac{6pq – p^2 – q^2}{4pq})