Trigonometrical Equations
easy

समीकरणों $\sin \theta  =  - \frac{1}{2}$ तथा $\tan \theta  = \frac{1}{{\sqrt 3 }}$ को सन्तुष्ट करने वाला $\theta $ का सर्वव्यापक मान है  

A

$n\pi + {( - 1)^n}\frac{\pi }{6}$

B

$n\pi + \frac{\pi }{6}$

C

$2n\pi \pm \frac{\pi }{6}$

D

इनमें से कोई नहीं

Solution

$\sin \theta  =  – \frac{1}{2} = \sin \left( { – \frac{\pi }{6}} \right) = \sin {\rm{ }}\left( {\pi  + \frac{\pi }{6}} \right)$

$\tan \theta  = \frac{1}{{\sqrt 3 }} = \tan \left( {\frac{\pi }{6}} \right) = \tan \left( {\pi  + \frac{\pi }{6}} \right) $

$\Rightarrow \theta  = \left( {\pi  + \frac{\pi }{6}} \right)$

अत: $\theta $ का व्यापक मान $2n\pi  + \frac{{7\pi }}{6}$ है।

 

Standard 11
Mathematics

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