# Need help with double angle question (1 Viewer)

#### dunce

##### New Member
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

##### π
$\bg_white LHS = tan(2x)cot(x)$

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

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

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

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

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

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

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

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

#### Drongoski

##### Well-Known Member
$\bg_white LHS = \frac {sin 2x}{cos 2x} \times \frac {cos x}{sin x} = \frac {2sinx\cdot cosx \cdot cos x}{cos 2x \times sin x} \\ \\ = \frac {2cos^2 x}{cos 2x} = \frac {cos 2x + 1}{cos 2x} = \frac {cos 2x}{cos 2x} + \frac {1}{cos 2x} = 1 + sec 2x$