Need help with double angle question (1 Viewer)

dunce · Aug 2, 2019

Hey everyone, could someone please show the working out for this question using double angles, not t-method? Thank you!

tan 2x cot x = 1 + sec 2x

Drdusk · Aug 2, 2019

$LHS = tan(2x)cot(x)$

$= \frac{sin(2x)}{cos(2x)}\times \frac{cos(x)}{sin(x)}$

$sin(2x) = sin(x+x) = sin(x)cos(x) + sin(x)cos(x) = 2sin(x)cos(x)\hspace{2mm} \text{(Double angle formula)}$

$cos(2x) = cos^2(x) - sin^2(x) \hspace{2mm} \text{(Double angle formula)}$

$\therefore LHS = \frac{2sin(x)cos(x)}{cos^2(x) - sin^2(x)}\times \frac{cos(x)}{sin(x)}$

$= \frac{2cos^2 (x)}{cos^2(x)-sin^2(x)}$

$= \frac{cos^2(x) - sin^2(x) + cos^2(x) + sin^2(x)}{cos^2(x) - sin^2(x)}$

$= \frac{cos^2(x) - sin^2(x)}{cos^2(x) - sin^2(x)} + \frac{cos^2(x) + sin^2(x)}{cos^2(x) - sin^2(x)}$

$= 1 + \frac{1}{cos(2x)} = 1 + sec(2x) = RHS$

Drongoski · Aug 2, 2019

Need help with double angle question (1 Viewer)

dunce

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Drdusk

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Drongoski

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