यदि $\tan \theta = \frac{a}{b}$ है,तो $\frac{{\sin \theta }}{{{{\cos }^8}\theta }} + \frac{{\cos \theta }}{{{{\sin }^8}\theta }} = $

  • A
    $\pm \frac{{{{({a^2} + {b^2})}^4}}}{{\sqrt {{a^2} + {b^2}} }}\left( {\frac{a}{{{b^8}}} + \frac{b}{{{a^8}}}} \right)$
  • B
    $\pm \frac{{{{({a^2} + {b^2})}^4}}}{{\sqrt {{a^2} + {b^2}} }}\left( {\frac{a}{{{b^8}}} - \frac{b}{{{a^8}}}} \right)$
  • C
    $\pm \frac{{{{({a^2} - {b^2})}^4}}}{{\sqrt {{a^2} + {b^2}} }}\left( {\frac{a}{{{b^8}}} + \frac{b}{{{a^8}}}} \right)$
  • D
    $\pm \frac{{{{({a^2} - {b^2})}^4}}}{{\sqrt {{a^2} - {b^2}} }}\left( {\frac{a}{{{b^8}}} - \frac{b}{{{a^8}}}} \right)$

Explore More

Similar Questions

यदि $\cos(\alpha + \beta) = \frac{3}{5}$,$\sin(\alpha - \beta) = \frac{5}{13}$ और $0 < \alpha, \beta < \frac{\pi}{4}$ है,तो $\tan(2\alpha)$ का मान ज्ञात कीजिए।

Difficult
View Solution

$(1+\cot A)^{2}+(1-\cot A)^{2}$ का मान क्या है?

$\frac{1}{4} [\sqrt{3} \cos 23^\circ - \sin 23^\circ] = $

$\frac{\sin 300^{\circ} \tan 240^{\circ} \sec \left(-420^{\circ}\right)}{\cot \left(-315^{\circ}\right) \cos \left(210^{\circ}\right) \operatorname{cosec}\left(-315^{\circ}\right)}$ का मान ज्ञात कीजिए।

यदि $x = a \operatorname{cosec}^{n} \theta$ और $y = b \cot^{n} \theta$ है,तो $\theta$ को विलुप्त करने पर क्या प्राप्त होगा?

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