# If tanx=-12/5, find ((cosx-1)(cosx+1))/((sinx-1)(sinx+1))?

Jun 29, 2016

$\frac{\left(\cos x - 1\right) \left(\cos x + 1\right)}{\left(\sin x - 1\right) \left(\sin x + 1\right)} = \frac{144}{25}$

#### Explanation:

As $\tan x = - \frac{12}{5}$, we have

$\cos x = \frac{1}{\sec} x = \frac{1}{\sqrt{{\sec}^{2} x}} = \frac{1}{\sqrt{1 + {\tan}^{2} x}}$

= $\frac{1}{\sqrt{1 + {\left(- \frac{12}{5}\right)}^{2}}} = \frac{1}{\sqrt{1 + \frac{144}{25}}} = \frac{1}{\sqrt{\frac{169}{25}}}$

= $\pm \frac{1}{\frac{13}{5}} = \pm \frac{5}{13}$ and

$\sin x = \sin \frac{x}{\cos} x \times \cos x = \tan x \times \cos x = - \frac{12}{5} \times \left(\pm \frac{5}{13}\right) = \pm \frac{12}{13}$

or ${\cos}^{2} x = \frac{25}{169}$ and ${\sin}^{2} x = \frac{144}{169}$

Hence $\frac{\left(\cos x - 1\right) \left(\cos x + 1\right)}{\left(\sin x - 1\right) \left(\sin x + 1\right)}$

= $\frac{{\cos}^{2} x - 1}{{\sin}^{2} x - 1} = \frac{\frac{25}{169} - 1}{\frac{144}{169} - 1}$

= $\frac{- \frac{144}{169}}{- \frac{25}{169}} = - \frac{144}{169} \times - \frac{169}{25} = \frac{144}{25}$

Jun 29, 2016

$\frac{144}{25}$

#### Explanation:

$\frac{\left(\cos x - 1\right) \left(\cos x + 1\right)}{\left(\sin x - 1\right) \left(\sin x + 1\right)}$

$= - \frac{\left(1 - \cos x\right) \left(1 + \cos x\right)}{-} \left(\left(1 - \sin x\right) \left(1 + \sin x\right)\right)$
$= \frac{1 - {\cos}^{2} x}{1 - {\sin}^{2} x}$

$= {\sin}^{2} \frac{x}{\cos} ^ 2 x = {\tan}^{2} x = {\left(- \frac{12}{5}\right)}^{2} = \frac{144}{25}$