Trigonometrical Equations
medium

यदि $\sqrt 2 \sec \theta  + \tan \theta  = 1,$ तो $\theta $ का व्यापक मान है  

A

$n\pi + \frac{{3\pi }}{4}$

B

$2n\pi + \frac{\pi }{4}$

C

$2n\pi - \frac{\pi }{4}$

D

$2n\pi \pm \frac{\pi }{4}$

Solution

$\sqrt 2 \sec \theta  + \tan \theta  = 1$

$\Rightarrow \frac{{\sqrt 2 }}{{\cos \theta }} + \frac{{\sin \theta }}{{\cos \theta }} = 1$

$ \Rightarrow $  $\sin \theta  – \cos \theta  =  – \sqrt 2 $

$\sqrt 2 $ से दोनों पक्षों में भाग देने पर,

$\frac{1}{{\sqrt 2 }}\sin \theta  – \frac{1}{{\sqrt 2 }}\cos \theta  =  – 1$

$ \Rightarrow $ $\frac{1}{{\sqrt 2 }}\cos \theta  – \frac{1}{{\sqrt 2 }}\sin \theta  = 1 $

$\Rightarrow \cos \,\left( {\theta  + \frac{\pi }{4}} \right) = \cos (0)$

$ \Rightarrow $ $\theta  + \frac{\pi }{4} = 2n\pi  \pm 0$

$\Rightarrow \theta  = 2n\pi  – \frac{\pi }{4}$.

Standard 11
Mathematics

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