$\frac{\cos 10^o + \sin 10^o}{\cos 10^o - \sin 10^o} = $

  • A
    $\tan 55^o$
  • B
    $\cot 55^o$
  • C
    $-\tan 35^o$
  • D
    $-\cot 35^o$

Explore More

Similar Questions

$1+\cos 10^{\circ}+\cos 20^{\circ}+\cos 30^{\circ}=$

If two acute angles $A$ and $B$ are such that $A \neq B$ and $\frac{x}{y}=\frac{\cos A}{\cos B}$,then $\frac{x \tan A-y \tan B}{x+y}=$

$\tan\, 20^{\circ} + \tan\, 40^{\circ} + \sqrt{3}\, \tan\, 20^{\circ} \tan\, 40^{\circ}$ is equal to

$\sin \left(\frac{\pi}{3}+x\right)-\cos \left(\frac{\pi}{6}+x\right) = $

If $\tan \alpha = \frac{m}{m + 1}$ and $\tan \beta = \frac{1}{2m + 1}$,then $\alpha + \beta = $

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