If sin x \(\frac{P - Q}{P + Q}\), where 0\(^o\) \(\leq\) x \(\leq\) 90\(^o\), find 1 - tan\(^2\)x

If sin x \(\frac{P - Q}{P + Q}\), where 0\(^o\) \(\leq\) x \(\leq\) 90\(^o\), find 1 - tan\(^2\)x

Answer Details

Let w be the adjacent side,

then w2 ${}^2$ = (P + q)2 ${}^2$ = 4pq so that w = 2pq−−√ $\sqrt{𝑝𝑞}$ 

Since tan x p−q2pq√ $\frac{𝑝-𝑞}{2\sqrt{𝑝𝑞}}$,

substituting;

1 - tan2 ${}^2$x = 1 - (p−q2pq√ $\frac{𝑝-𝑞}{2\sqrt{𝑝𝑞}}$)2 ${}^2$ 

= 1 - P2−2pq+q24pq $\frac{{𝑃}^2-2𝑝𝑞+{𝑞}^2}{4𝑝𝑞}$

= 4pq−p2+2pq−q24pq $\frac{4𝑝𝑞-{𝑝}^2+2𝑝𝑞-{𝑞}^2}{4𝑝𝑞}$

Therefore,

1 - tan2 ${}^2$x = 6pq−p2−q24pq