If sec theta = frac{13}{12} and theta lies in | vedclass

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Question

(A) Given $\sec 	heta = \frac{13}{12}$,therefore $\cos 	heta = \frac{12}{13}$.Since $	heta$ lies in the $4^{	ext{th}}$ quadrant,$\sin 	heta$ is n

Vedclass · Accepted Answer

$\frac{-5}{13}$

Vedclass · Answer

(A) Given $\sec 	heta = \frac{13}{12}$,therefore $\cos 	heta = \frac{12}{13}$.Since $	heta$ lies in the $4^{	ext{th}}$ quadrant,$\sin 	heta$ is negative.Using $\sin^2 	heta + \cos^2 	heta = 1$,we get $\sin 	heta = -\sqrt{1 - (\frac{12}{13})^2} = -\sqrt{1 - \frac{144}{169}} = -\sqrt{\frac{25}{169}} = -\frac{5}{13}$.Now,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{-5/13}{12/13} = -\frac{5}{12}$ and $\operatorname{cosec} 	heta = \frac{1}{\sin 	heta} = -\frac{13}{5}$.Substituting these values into the expression: $	an 	heta 	imes \operatorname{cosec} 	heta 	imes \sin 	heta 	imes \cos 	heta = (-\frac{5}{12}) 	imes (-\frac{13}{5}) 	imes (-\frac{5}{13}) 	imes (\frac{12}{13})$.$= (\frac{5}{12} 	imes \frac{13}{5}) 	imes (-\frac{5}{13} 	imes \frac{12}{13}) = (\frac{13}{12}) 	imes (-\frac{60}{169}) = -\frac{5}{13}$.

If $\sec \theta = \frac{13}{12}$ and $\theta$ lies in the $4^{\text{th}}$ quadrant,then $\tan \theta \times \operatorname{cosec} \theta \times \sin \theta \times \cos \theta = $

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