If $\frac{3\pi}{4} < \alpha < \pi,$ then $\sqrt{\csc^2 \alpha + 2\cot \alpha}$ is equal to
- A$1 + \cot \alpha$
- B$1 - \cot \alpha$
- C$-1 - \cot \alpha$
- D$-1 + \cot \alpha$
Explore More
Similar Questions
The value of $\tan 20^{\circ} + \tan 40^{\circ} + \tan 60^{\circ} + \cdots + \tan 180^{\circ}$ is
Medium
View SolutionThe value of $6(\sin^6 \theta + \cos^6 \theta) - 9(\sin^4 \theta + \cos^4 \theta) + 4$ is
Medium
View SolutionIf $5 \tan \theta = 4$,then $\frac{5 \sin \theta - 3 \cos \theta}{5 \sin \theta + 2 \cos \theta} = $
Easy
View SolutionIf $\alpha$ and $\beta$ are solutions of $\sin^2 x + a \sin x + b = 0$ as well as $\cos^2 x + c \cos x + d = 0$,then $\sin(\alpha + \beta)$ is equal to
If $\frac{\sec \theta+\tan \theta}{\sec \theta-\tan \theta}=2 \frac{51}{79}$,then the value of $\sin \theta$ is
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo