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Question

Prove that $$\cfrac{\sec\theta+\tan\theta-1}{\tan\theta-\sec\theta+1}$$ =$$ \cfrac{\cos\theta}{1-\sin\theta}$$


Solution

$$\cfrac{\sec\theta+\tan\theta-1}{\tan\theta-\sec\theta+1}$$
$$\Rightarrow \cfrac{\tan\theta+\sec\theta-(\sec^{2}\theta-\tan^{2}\theta)}{\tan\theta-\sec\theta+1}$$
$$\Rightarrow \cfrac{(\tan\theta+\sec\theta)-(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)}{\tan\theta-\sec\theta+1}$$
$$\Rightarrow \cfrac{(\tan\theta+\sec\theta)-(1-\sec\theta+\tan\theta)}{\tan\theta-\sec\theta+1}$$
$$\Rightarrow \sec\theta+\tan\theta$$
$$\Rightarrow \cfrac{1}{\cos\theta}+\cfrac{\sin\theta}{\tan\theta}$$
$$\Rightarrow \cfrac{1+\sin\theta}{\cos\theta}$$
$$\Rightarrow \cfrac{1+\sin\theta}{\cos\theta} \times \cfrac{1-\sin\theta}{1-\sin\theta}$$
$$\Rightarrow \cfrac{1-\sin^{2}\theta}{\cos\theta(1-\sin\theta)}$$
$$\Rightarrow \cfrac{\cos^{2}\theta}{\cos\theta(1-\sin\theta)}$$
$$\Rightarrow \cfrac{\cos\theta}{1-\sin\theta}$$ (Hence proved)





Mathematics

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