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Question

Prove that $$\frac{{1 + \cos \theta }}{{1 - \cos \theta }} = \frac{{{{\tan }^2}\theta }}{{{{\left( {\sec \theta  - 1} \right)}^2}}}$$


Solution

$$R.H.S.$$
$$\dfrac{1+\cos\theta}{1-\cos\theta}$$
Divide and multiply above expression by $$1-\cos\theta$$ we get
$$\bigg(\dfrac{1+\cos\theta}{1-\cos\theta}\bigg)\bigg(\dfrac{1-\cos\theta}{1-\cos\theta}\bigg)$$
$$\bigg(\dfrac{1-\cos^{2}\theta}{(1-\cos\theta)^{2}}\bigg)$$        $$\because (a+b)(a-b)=(a^{2}-b^{2}) $$
$$\bigg(\dfrac{\sin^{2}\theta}{(1-\cos\theta)^{2}}\bigg)$$ 
divide numerator and denominator by $$cos^{2}\theta$$  we get
$$\bigg(\dfrac{\dfrac{\sin^{2}\theta}{\cos^{2}\theta}}{\dfrac{(1-\cos\theta)^{2}}{\cos^{2}\theta}}\bigg)$$

$$\dfrac{tan^{2}\theta}{\bigg(\dfrac{1}{\cos\theta}-1\bigg)^{2}}$$        $$\because \dfrac{\sin\theta}{\cos\theta}=\tan\theta$$

$$\dfrac{\tan^{2}\theta}{(\sec\theta-1)^{2}}=L.H.S$$
Hence proved.

Mathematics

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