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Thus L.H.S = R.H.S that is 2/√3cosx + sinx = sec(Π/6-x) is proved
We have to prove that
2/√3cosx + sinx = sec(Π/6-x)
To prove this we will solve the right-hand side of the equation which is
R.H.S = sec(Π/6-x)
= 1/cos(Π/6-x)
[As secƟ = 1/cosƟ)
= 1/[cos Π/6cosx + sin Π/6sinx]
[As cos (X-Y) = cosXcosY + sinXsinY , which is a trigonometry identity where X = Π/6 and Y = x]
= 1/[√3/2cosx + 1/2sinx]
= 1/(√3cosx + sinx]/2
= 2/√3cosx + sinx
R.H.S = L.H.S
Hence 2/√3cosx + sinx = sec(Π/6-x) is proved
Learn more about trigonometry here : brainly.com/question/7331447
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Answer is 11w-6
Add all the like terms. So all with the w’s which is how I got 11 and then 6 stays the same hence the -6
Answer:
Midpoint
Step-by-step explanation:
The midpoint is the point on the segment halfway between the endpoints. It may be the case that the midpoint of a segment can be found simply by counting. If the segment is horizontal or vertical, you can find the midpoint by dividing the length of the segment by 2 and counting that value from either of the endpoints.
