If $ an heta = \frac{3}{4}$, find the value of $\left(\frac{1 - \cos^2 heta}{1 + \cos^2 heta} ight)$.

Question

If $	an 	heta = \frac{3}{4}$, find the value of $\left(\frac{1 - \cos^2 	heta}{1 + \cos^2 	heta}ight)$.

Accepted Answer

Step 1: $\sec^2 	heta = 1 + \frac{9}{16} = \frac{25}{16}$. $	herefore \cos^2 	heta = \frac{16}{25}$
Step 2: Hence $\frac{1 - \cos^2 	heta}{1 + \cos^2 	heta} = \frac{1 - \frac{16}{25}}{1 + \frac{16}{25}} = \frac{\frac{9}{25}}{\frac{41}{25}} = \frac{9}{41}$

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