If $\sec \,\,\theta \,\,\tan \,\,\theta = \sqrt 2 $, then $\theta = $

$2\;n\;\pi \,\, \pm \frac{\pi }{3},\,\,n \in Z$

$n\;\pi \,\, \pm \,\,\frac{{2\;\pi }}{3},\,\,n \in Z$

$n\;\pi \,\, - \,\,\frac{\pi }{4},\,\,n \in Z$

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

\sec \,\,\theta \,\,\tan \,\,\theta = \sqrt 2

, then

\theta =

View answer and explanation

Correct option: A

n\;\pi \,\, + \,\,{\left( { - 1} \right)^n}\frac{\pi }{4},\,\,n \in Z

Option A (

n\\;\pi \\, + \\,{\left( { - 1} \right)^n}\frac{\pi }{4},\\,n \in Z

) is correct because it represents the general solution for

\theta

derived from the given equation

\sec \theta \tan \theta = \sqrt 2

Practice All Chapter MCQs →

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