For any real values of theta,sqrt{frac{sec th | vedclass

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Question

(C) We start with the expression $\sqrt{\frac{\sec 	heta - 1}{\sec 	heta + 1}}$.Converting $\sec 	heta$ to $\frac{1}{\cos 	heta}$,we get $\sqrt{\f

Vedclass · Accepted Answer

$\operatorname{cosec} 	heta - \cot 	heta$

Vedclass · Answer

(C) We start with the expression $\sqrt{\frac{\sec 	heta - 1}{\sec 	heta + 1}}$.Converting $\sec 	heta$ to $\frac{1}{\cos 	heta}$,we get $\sqrt{\frac{\frac{1}{\cos 	heta} - 1}{\frac{1}{\cos 	heta} + 1}} = \sqrt{\frac{1 - \cos 	heta}{1 + \cos 	heta}}$.Multiplying the numerator and denominator by $(1 - \cos 	heta)$,we get $\sqrt{\frac{(1 - \cos 	heta)^2}{1 - \cos^2 	heta}}$.Using the identity $1 - \cos^2 	heta = \sin^2 	heta$,the expression becomes $\sqrt{\frac{(1 - \cos 	heta)^2}{\sin^2 	heta}} = \frac{1 - \cos 	heta}{\sin 	heta}$.This simplifies to $\frac{1}{\sin 	heta} - \frac{\cos 	heta}{\sin 	heta} = \operatorname{cosec} 	heta - \cot 	heta$.

For any real values of $\theta$,$\sqrt{\frac{\sec \theta-1}{\sec \theta+1}} = ?$

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