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Question Number 63060 by rajesh4661kumar@gamil.com last updated on 28/Jun/19

Answered by Hope last updated on 28/Jun/19

$$\sqrt{\frac{{(1+cos\theta )}^2}{{sin}^2\theta }}$$$$=\mid \frac{1+cos\theta }{sin\theta }\mid$$$$=\frac{\mid 1+cos\theta \mid }{\mid sin\theta \mid }$$$$when\pi >\theta >0sosin\theta =+ve$$$$cos\theta <0but\mid cos\theta \mid <1\mid 1+cos\theta \mid =1+cos\theta$$$$so\frac{\mid 1+cos\theta \mid }{\mid sin\theta \mid }$$$$=\frac{1+cos\theta }{sin\theta }=cosec\theta +cot\theta$$$$\mathit{b}\mathit{u}\mathit{t}\mathit{w}\mathit{h}\mathit{e}\mathit{n}2\pi >\theta >\pi$$$$=\frac{\mid 1+cos\theta \mid }{\mid sin\theta \mid }$$$$=\frac{1+cos\theta }{-sin\theta }$$$$=-cosec\theta -cot\theta$$$$$$

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