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$\tan ^{ }{ \theta }$ in terms of $\cos ^{ }{ \theta } $ in quadrant III

I suppose that using the second pythagorean identity: $\tan ^{ 2 }{ \theta } +1=\sec ^{ 2 }{ \theta } $

I have a general idea of how to manipulate this equation... but I am stuck in terms of arriving at the correct answer. This is as far as I have gotten:

$$\tan ^{ 2 }{ \theta } =\sec ^{ 2 }{ \theta } -1$$ $$\tan ^{ 2 }{ \theta } =\frac { 1 }{ \cos ^{ 2 }{ \theta } } -1$$ $$\tan ^{ 2 }{ \theta } =\frac { 1 }{ \cos ^{ 2 }{ \theta } } -(\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } )$$

I feel very lost going from here. I probably didn't even head in the right direction to begin with. I would prefer a hint to set me on the right path rather than a direct answer.

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1 Answer

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In general $$\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}=\frac{1-\cos^2\theta}{\cos^2\theta}$$

Now in the $III$ Quadrant $\tan\theta\ge0$

So, $$\tan\theta=+\sqrt{\frac{1-\cos^2\theta}{\cos^2\theta}}$$

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