sqrt{frac{1+cos theta}{1-cos theta}}+sqrt{fra | vedclass

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(D) Let the expression be $E = \sqrt{\frac{1+\cos 	heta}{1-\cos 	heta}} + \sqrt{\frac{1-\cos 	heta}{1+\cos 	heta}}$.Taking the common denominator,

Vedclass · Accepted Answer

$\frac{2}{|\sin 	heta|}$

Vedclass · Answer

(D) Let the expression be $E = \sqrt{\frac{1+\cos 	heta}{1-\cos 	heta}} + \sqrt{\frac{1-\cos 	heta}{1+\cos 	heta}}$.Taking the common denominator,we get:$E = \frac{\sqrt{1+\cos 	heta} \cdot \sqrt{1+\cos 	heta} + \sqrt{1-\cos 	heta} \cdot \sqrt{1-\cos 	heta}}{\sqrt{1-\cos 	heta} \cdot \sqrt{1+\cos 	heta}}$$E = \frac{(1+\cos 	heta) + (1-\cos 	heta)}{\sqrt{(1-\cos 	heta)(1+\cos 	heta)}}$$E = \frac{2}{\sqrt{1-\cos^2 	heta}}$Since $1-\cos^2 	heta = \sin^2 	heta$,we have $\sqrt{\sin^2 	heta} = |\sin 	heta|$.Therefore,$E = \frac{2}{|\sin 	heta|}$.

$\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}+\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=$

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