$0 \leq x \leq \frac{\pi}{4}$ के लिए समीकरण $32^{\tan^{2} x} + 32^{\sec^{2} x} = 81$ के हलों की संख्या क्या है?

  • A
    $3$
  • B
    $1$
  • C
    $0$
  • D
    $2$

Explore More

Similar Questions

यदि $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}$ है,तो
$(A) \tan ^2 x=\frac{2}{3}$ $(B) \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}$
$(C) \tan ^2 x=\frac{1}{3}$ $(D) \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{2}{125}$

यदि $\sin 10^{\circ} \sin 50^{\circ} \sin 60^{\circ} \sin 70^{\circ} = m$ और $\tan 20^{\circ} \tan 40^{\circ} \tan 60^{\circ} \tan 80^{\circ} = n$ है,तो $\frac{n}{m}$ का मान ज्ञात कीजिए।

यदि $a \sin^2 x + b \cos^2 x = c$,$b \sin^2 y + a \cos^2 y = d$ और $a \tan x = b \tan y$ है,तो $\frac{a^2}{b^2}$ का मान ज्ञात कीजिए।

Difficult
View Solution

$\tan 81^{\circ} - \tan 63^{\circ} - \tan 27^{\circ} + \tan 9^{\circ}$ का मान है

यदि $0 \leq A, B \leq \frac{\pi}{4}$ और $\cot A + \cot B + \tan A + \tan B = \cot A \cot B - \tan A \tan B$ है,तो $\sin(A + B) = $

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