यदि $\sec x = \frac{25}{24}$ और $x$ प्रथम चतुर्थांश में स्थित है,तो $\sin \frac{x}{2} + \cos \frac{x}{2} =$

  • A
    $\frac{6}{5 \sqrt{2}}$
  • B
    $\frac{8}{5 \sqrt{2}}$
  • C
    $\frac{7}{5 \sqrt{2}}$
  • D
    $\frac{1}{5 \sqrt{2}}$

Explore More

Similar Questions

$\tan 3A \cdot \tan 2A \cdot \tan A = $

$\tan \left(\frac{3 \pi}{16}\right)+\cot \left(\frac{3 \pi}{16}\right)=$

यदि $\frac{1}{2}\left(\tan \left(\frac{\pi}{24}\right)+\cot \left(\frac{\pi}{24}\right)\right)=\sqrt{a^2+a}+\sqrt{a}$ है,तो $a=$

$(1+\sec 2\theta)(1+\sec 4\theta) = $

कुछ $a, b, c \in R$ के लिए,यदि $\sin 5 \theta = a \cos^4 \theta \sin \theta + b \cos^2 \theta \sin^3 \theta + c \sin^5 \theta$ है,तो $abc =$

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