If sin x \(\frac{P - Q}{P + Q}\), where 0\(^o\) \(\leq\) x \(\leq\) 90\(^o\), find...
FURTHER MATHEMATICS
WAEC 2019
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 \(\frac{p - q}{2\sqrt{pq}}\),
substituting; 1 - tan\(^2\)x = 1 - (\(\frac{p - q}{2\sqrt{pq}}\))\(^2\)
= 1 - \(\frac{P^2 - 2pq + q^2}{4pq}\)
= \(\frac{4pq - p^2 + 2pq - q^2}{4pq}\)
Therefore, 1 - tan\(^2\)x = \(\frac{6pq - p^2 - q^2}{4pq}\)
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