If sin theta = frac{24}{25} and theta lies in | vedclass

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

Question

(C) Given $\sin 	heta = \frac{24}{25}$.Since $	heta$ lies in the second quadrant,$\cos 	heta$ and $	an 	heta$ are negative.Using the identity $\c

Vedclass · Accepted Answer

$-7$

Vedclass · Answer

(C) Given $\sin 	heta = \frac{24}{25}$.Since $	heta$ lies in the second quadrant,$\cos 	heta$ and $	an 	heta$ are negative.Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$,we get $\cos^2 	heta = 1 - (\frac{24}{25})^2 = 1 - \frac{576}{625} = \frac{49}{625}$.Since $	heta$ is in the second quadrant,$\cos 	heta = -\sqrt{\frac{49}{625}} = -\frac{7}{25}$.Then,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{24/25}{-7/25} = -\frac{24}{7}$.Also,$\sec 	heta = \frac{1}{\cos 	heta} = -\frac{25}{7}$.Therefore,$\sec 	heta + 	an 	heta = -\frac{25}{7} + (-\frac{24}{7}) = -\frac{49}{7} = -7$.

If $\sin \theta = \frac{24}{25}$ and $\theta$ lies in the second quadrant,then $\sec \theta + \tan \theta = $

Similar Questions

The maximum value of $4\sin^2 x + 3\cos^2 x$ is

If $\tan A + \cot A = 4,$ then ${\tan ^4}A + {\cot ^4}A$ is equal to

The value of $\csc \frac{\pi}{18} - \sqrt{3} \sec \frac{\pi}{18}$ is a

If $\sin \theta + \cos \theta = m$ and $\sec \theta + \csc \theta = n$,then $n(m + 1)(m - 1) = $

The value of $\cot\, x + \cot\, (60^\circ + x) + \cot\, (120^\circ + x)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module