Question 43
$$\frac{\left(1+\cos\theta\right)^2+\sin^2\theta}{\left(\text{coec}^2\theta-1\right)\sin^2\theta}=$$
Solution
$$\frac{\left(1+\cos\theta\right)^2+\sin^2\theta}{\left(\text{coec}^2\theta-1\right)\sin^2\theta}$$
=$$\frac{1 + \cos^2\theta + 2\cos\theta +\sin^2\theta}{1 -\sin^2\theta}$$
=$$\frac{2 + 2\cos\theta }{\cos^2\theta}$$
=$$2(\sec^2\theta+ \sec\theta)$$
=$$2\sec\theta(1 + \sec\theta)$$
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