If $3 \sin \alpha = 5 \sin \beta$,then $\tan \left(\frac{\alpha + \beta}{2}\right) \div \tan \left(\frac{\alpha - \beta}{2}\right) = $

  • A
    $1$
  • B
    $2$
  • C
    $3$
  • D
    $4$

Explore More

Similar Questions

Let $\alpha$ and $\beta$ be real numbers such that $-\frac{\pi}{4} < \beta < 0 < \alpha < \frac{\pi}{4}$. If $\sin (\alpha+\beta) = \frac{1}{3}$ and $\cos (\alpha-\beta) = \frac{2}{3}$,then the greatest integer less than or equal to $\left(\frac{\sin \alpha}{\cos \beta} + \frac{\cos \beta}{\sin \alpha} + \frac{\cos \alpha}{\sin \beta} + \frac{\sin \beta}{\cos \alpha}\right)^2$ is:

Let $[x]$ denote the largest integer $\leq x$. If the number of solutions of $\sin x \sqrt{4 \cos ^2 x} = \frac{2+x-[x]}{1-x+[x]}$ is $k$,then for $x \in \left[\frac{\pi}{4}, \frac{\pi}{3}\right]$,the value of $k^{\tan^2 x}$

If $\sin 2\theta + \sin 2\phi = \frac{1}{2}$ and $\cos 2\theta + \cos 2\phi = \frac{3}{2}$,then $\cos^2(\theta - \phi) =$

If $\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=x$,then $\frac{1-\cos \alpha-\sin \alpha}{\cos \alpha}=$

If the roots of $x^3 + ax^2 + bx + c = 0$ are the cosines of the angles of an acute triangle,then the value of $a^2 - 2b - 2c$ is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo