For all theta, the value of sqrt{frac{1+sin t | vedclass

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(D) To simplify the expression $\sqrt{\frac{1+\sin 	heta}{1-\sin 	heta}},$ we multiply the numerator and denominator inside the square root by $(1+\

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$\sec 	heta + 	an 	heta$

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(D) To simplify the expression $\sqrt{\frac{1+\sin 	heta}{1-\sin 	heta}},$ we multiply the numerator and denominator inside the square root by $(1+\sin 	heta).$$= \sqrt{\frac{(1+\sin 	heta)(1+\sin 	heta)}{(1-\sin 	heta)(1+\sin 	heta)}}$$= \sqrt{\frac{(1+\sin 	heta)^2}{1-\sin^2 	heta}}$Since $1-\sin^2 	heta = \cos^2 	heta,$$= \sqrt{\frac{(1+\sin 	heta)^2}{\cos^2 	heta}}$$= \frac{1+\sin 	heta}{\cos 	heta}$$= \frac{1}{\cos 	heta} + \frac{\sin 	heta}{\cos 	heta}$$= \sec 	heta + 	an 	heta$

For all $\theta,$ the value of $\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}$ is:

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