JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Easy

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Application +4 | -1

\cot \,\,x + \mathrm{cose}c\;\,\,x = \sqrt 3

, then the principle value of

\left( {x - \frac{\pi }{6}} \right)

View answer and explanation

Correct option: D

\frac{\pi }{6}

Option D (

\frac{\pi }{6}

) is correct because solving the given trigonometric equation

\cot x + \csc x = \sqrt{3}

leads to

x = 2n\pi + \frac{\pi}{3}

, and substituting into

x - \frac{\pi}{6}

gives

\frac{\pi}{6}

for

n=0

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