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2019 WAEC Further Mathematics Theory If sin x \(\frac{P - Q}{P + Q}\), where 0\(^o\) \(\leq\) x \(\leq\) 90\(^o\), find 1...

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

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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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