If sin theta = frac{-12}{13},cos phi = frac{- | vedclass

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

Question

(D) Given $\sin 	heta = \frac{-12}{13}$ and $	heta$ is in the third quadrant,$\cos 	heta = -\sqrt{1 - \sin^2 	heta} = -\sqrt{1 - (\frac{-12}{13})^

Vedclass · Accepted Answer

$\frac{33}{56}$

Vedclass · Answer

(D) Given $\sin 	heta = \frac{-12}{13}$ and $	heta$ is in the third quadrant,$\cos 	heta = -\sqrt{1 - \sin^2 	heta} = -\sqrt{1 - (\frac{-12}{13})^2} = -\sqrt{\frac{25}{169}} = -\frac{5}{13}$.Thus,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{-12/13}{-5/13} = \frac{12}{5}$.Given $\cos \phi = \frac{-4}{5}$ and $\phi$ is in the third quadrant,$\sin \phi = -\sqrt{1 - \cos^2 \phi} = -\sqrt{1 - (\frac{-4}{5})^2} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$.Thus,$	an \phi = \frac{\sin \phi}{\cos \phi} = \frac{-3/5}{-4/5} = \frac{3}{4}$.Using the formula $	an(	heta - \phi) = \frac{	an 	heta - 	an \phi}{1 + 	an 	heta 	an \phi}$:$	an(	heta - \phi) = \frac{12/5 - 3/4}{1 + (12/5)(3/4)} = \frac{(48 - 15)/20}{1 + 36/20} = \frac{33/20}{56/20} = \frac{33}{56}$.

If $\sin \theta = \frac{-12}{13}$,$\cos \phi = \frac{-4}{5}$ and $\theta, \phi$ lie in the third quadrant,then $\tan(\theta - \phi) =$

Similar Questions

Prove that $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \left(\frac{x-y}{2}\right)$

$\frac{\tan 80^{\circ}-\tan 10^{\circ}}{\tan 70^{\circ}}$ is equal to

If $\cos (A + B) = \alpha \cos A \cos B + \beta \sin A \sin B,$ then $(\alpha, \beta) =$

Prove that $\sin (n+1) x \sin (n+2) x + \cos (n+1) x \cos (n+2) x = \cos x$.

If $\cot \alpha = 1$ and $\sec \beta = -\frac{5}{3}$,where $\pi < \alpha < \frac{3\pi}{2}$ and $\frac{\pi}{2} < \beta < \pi$,then the value of $\tan(\alpha + \beta)$ and the quadrant in which $\alpha + \beta$ lies,respectively,are

Vedclass Test Series

Exam Paper Generator

Online Exam Module