$2{\tan ^{ - 1}}\frac{1}{3} + {\tan ^{ - 1}}\frac{1}{2} = $

  • A
    $90^o$
  • B
    $60^o$
  • C
    $45^o$
  • D
    $\tan ^{ - 1}2$

Explore More

Similar Questions

$\tan^{-1}4x + \tan^{-1}6x = \frac{\pi}{6}$ के हलों की संख्या ज्ञात कीजिए,जहाँ $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$ है।

${(\sin ^{ - 1}}x)^3 + {(\cos ^{ - 1}}x)^3$ का अधिकतम और न्यूनतम मान ज्ञात कीजिए।

Difficult
View Solution

यदि $\alpha$ और $\beta$ प्रथम चतुर्थांश में ऐसे कोण हैं कि $\tan \alpha = \frac{1}{7}$ और $\sin \beta = \frac{1}{\sqrt{10}}$,तो $\alpha + 2\beta =$ ($^{\circ}$ में)

यदि ${x^2} + {y^2} + {z^2} = {r^2}$ है,तो ${\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{zx}}{{yr}}} \right) = $

यदि $\theta = \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{13}\right) + \tan^{-1}\left(\frac{1}{21}\right) + \tan^{-1}\left(\frac{1}{31}\right)$,तो $\tan \theta =$

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