frac{sec theta-1}{sec theta+1} = ldots | vedclass

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(D) We are given the expression $\frac{\sec 	heta-1}{\sec 	heta+1}$.Since $\sec 	heta = \frac{1}{\cos 	heta}$,we can multiply the numerator and th

Vedclass · Accepted Answer

$\frac{1-\cos 	heta}{1+\cos 	heta}$

Vedclass · Answer

(D) We are given the expression $\frac{\sec 	heta-1}{\sec 	heta+1}$.Since $\sec 	heta = \frac{1}{\cos 	heta}$,we can multiply the numerator and the denominator by $\cos 	heta$:$\frac{\sec 	heta-1}{\sec 	heta+1} = \frac{(\sec 	heta-1) \cdot \cos 	heta}{(\sec 	heta+1) \cdot \cos 	heta}$$= \frac{\sec 	heta \cdot \cos 	heta - \cos 	heta}{\sec 	heta \cdot \cos 	heta + \cos 	heta}$$= \frac{1 - \cos 	heta}{1 + \cos 	heta}$ (since $\sec 	heta \cdot \cos 	heta = 1$)Thus,the correct option is $D$.

Part $I$	Part $II$
$1.$ $\cos(90^\circ - \theta)$	$a.$ $\sec \theta$
$2.$ $\cot(90^\circ - \theta)$	$b.$ $\sin \theta$
$3.$ $\operatorname{cosec}(90^\circ - \theta)$	$c.$ $1$
	$d.$ $\tan \theta$

$\frac{\sec \theta-1}{\sec \theta+1} = \ldots$

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Which of the following groups truly matches the data of Part $I$ with the data of Part $II$?

Part $I$ Part $II$

$1.$ $\cos(90^\circ - \theta)$ $a.$ $\sec \theta$

$2.$ $\cot(90^\circ - \theta)$ $b.$ $\sin \theta$

$3.$ $\operatorname{cosec}(90^\circ - \theta)$ $c.$ $1$

$d.$ $\tan \theta$

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