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Prove the following trigonometric identity: 2/ (√3 cos x + sin x) = sec(π/6 - x)
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\[\dfrac{2}{\sqrt3 cos x+sin x}=sec(\pi/6 -x)\] right?
Yes. I have no idea how to do that on this site, but yes
let go from the right hand side \[sec (\pi/6-x)=\dfrac{1}{cos(\pi/6-x)}\]
Now, manipulate a little bit the denominator cos (pi/6 -x) = cos (pi/6) cos x+ sin(pi/6) sinx cos pi/6= sqrt3/2 sin pi/6 = 1/2 just plug on them in , simplify and you get the left hand side
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That looks about right. Thank you!
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