First slide

Compound angles

Question

If $\cot^{2} x = \cot (x - y) \cot (x - z)$ ,then cot2x is equal to (where $x \neq \pm π / 4$ )

Moderate

A

$\frac{1}{2} (\tan y + \tan z)$
B

$\frac{1}{2} (\cot y + \cot z)$
C

$\frac{1}{2} (\sin y + \sin z)$
D

none of these

Solution

$\cot^{2} x = \cot (x - y) \cot (x - z)$
$\begin{array}{l} or \cot^{2} x = (\frac{\cot xcot y + 1}{\cot y - \cot x}) (\frac{\cot xcot z + 1}{\cot z - \cot x}) \\ or \cot^{2} xcot ycot z - \cot^{3} xcot y - \cot^{3} xcot z + \cot^{4} x \\ = \cot^{2} xcot ycot z + \cot xcot y + \cot xcot z + 1 \\ or \cot^{3} x (\cot y + \cot z) + \cot x (\cot y + \cot z) + 1 - \cot^{4} x = 0 \\ or \cot x (\cot y + \cot z) (1 + \cot^{2} x) + (1 - \cot^{2} x) (1 + \cot^{2} x) = 0 \\ or [\cot x (\cot y + \cot z) + (1 - \cot^{2} x)] = 0 \\ or \frac{\cot^{2} x - 1}{2 \cot x} = \frac{1}{2} (\cot y + \cot z) \end{array}$

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