Which of the following relations is possible?

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

Question

(B) For any real angle $	heta$:$1$. The range of $\sin 	heta$ and $\cos 	heta$ is $[-1, 1]$. Thus,$\sin 	heta = \frac{5}{3} > 1$ is impossible.$2$

Vedclass · Accepted Answer

$	an 	heta = 1002$

Vedclass · Answer

(B) For any real angle $	heta$:$1$. The range of $\sin 	heta$ and $\cos 	heta$ is $[-1, 1]$. Thus,$\sin 	heta = \frac{5}{3} > 1$ is impossible.$2$. The range of $\sec 	heta$ is $(-\infty, -1] \cup [1, \infty)$. Thus,$\sec 	heta = \frac{1}{2}$ is impossible because $\frac{1}{2} $3$. For $\cos 	heta = \frac{1 + p^2}{1 - p^2}$,let $p = 2$. Then $\cos 	heta = \frac{1+4}{1-4} = \frac{5}{-3} = -1.66$,which is less than $-1$,so it is impossible.$4$. The range of $	an 	heta$ is $(-\infty, \infty)$. Therefore,$	an 	heta = 1002$ is possible.

Similar Questions

$\sin 50^\circ - \sin 70^\circ + \sin 10^\circ = $

If $y = \log_e \tan \left(\frac{\pi}{4} + \frac{x}{2}\right)$,then $\tanh \left(\frac{y}{2}\right) = $

If $\sin A = \frac{-7}{25}$,$\cos B = \frac{8}{17}$,$A$ does not lie in the $3^{\text{rd}}$ quadrant and $B$ does not lie in the $1^{\text{st}}$ quadrant,then $8 \tan A - 5 \cot B =$

The value of $\text{sech}(i\pi)$ is

$\frac{\sin ^2(-160^{\circ})}{\sin ^2 70^{\circ}} + \frac{\sin (180^{\circ}-\theta)}{\sin \theta} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module