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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 = $

Similar Questions

$\sin (\pi + \theta )\sin (\pi - \theta )\csc^2 \theta = $

If $\operatorname{cosec} \theta - \cot \theta = 2017$,then the quadrant in which $\theta$ lies is

The value of $\cos A - \sin A$ when $A = \frac{5\pi}{4}$ is

$\sin ^2 \frac{2 \pi}{3}+\cos ^2 \frac{5 \pi}{6}-\tan ^2 \frac{3 \pi}{4}=$

Find the degree measure corresponding to the following radian measure (Use $\pi = \frac{22}{7}$):
$-4$ radians

Vedclass Test Series

Exam Paper Generator

Online Exam Module