Question

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The general solution of equation 3tan(theta-15°) = tan(theta+15°)




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Answer 1

$\large{\mathbf{HELLO \: \: DEAR,}}$

$\mathit{GIVEN \: \: THAT:-}$

$\mathit{3tan(\theta - 15\degree) = tan(\theta + 15\degree)} \\ \\ \mathit{\frac{3}{1} = \frac{tan(\theta + 15\degree)}{tan(\theta - 15\degree)}}$

$\mathit{\frac{3 + 1}{3 - 1} = \frac{tan(\theta + 15\degree) + tan(\theta - 15\degree)}{tan(\theta + 15\degree) - tan(\theta - 15\degree)}}$

$\frac{4}{2} = \frac{\frac{sin(\theta + 15\degree)}{cos(\theta + 15\degree)} + \frac{sin(\theta - 15\degree)}{cos(\theta - 15\degree)}}{\frac{sin(\theta + 15\degree)}{cos(\theta + 15\degree)} - \frac{sin(\theta - 15\degree)}{cos(\theta - 15\degree)}}$

$\mathit{2= \frac{sin(\theta + 15\degree)*cos(\theta - 15\degree) + sin(\theta - 15\degree)*cos(\theta + 15\degree)}{sin(\theta + 15\degree)*cos(\theta - 15\degree) - sin(\theta - 15\degree)*cos(\theta + 15\degree)}}$

$\mathit{2 = \frac{sin(\theta + 15\degree + \theta - 15\degree)}{sin(\theta + 15\degree - \theta + 15\degree)}}$

$\mathit{2 = \frac{sin2\theta}{sin30\degree}}$

$\mathit{2 * \frac{1}{2} = sin{2\theta}}$

$\mathit{sin\frac{\pi}{2} = sin2\theta}$

$\mathit{WE \:\:KNOW \:\:THAT:-}$

$\mathit{GENERAL\:\: SOLUTION\: \: FOR \:\: sin\theta = sin\alpha}$

$\theta = n\pi + ( - 1)^{n} \alpha$
$\mathit{ \underline{NOW,}} \\ \\ \mathit{2\theta = {n\pi} + (-1)^n \frac{\pi}{2}}\\ \\ \mathit{\theta = \frac{n\pi}{2} + (-1)^n\frac{\pi}{4}}$

$\large{\mathit{\underline{I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}}}$