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
#SPJ9
For the average to be 90,
(99 + 93 + x) / 3 = 90
or 192 + x = 270
or x = 270 - 192 = 78
Jim must score at least 78 in the third test.
Find that median 169, 72, 161, 322, 56, 145, 64,64, 80, 288, 95, 97, 209,209,48
Alexus [3.1K]
Answer:
97
Step-by-step explanation:
The number in the middle is 97
Answer:
The end behavior of a function f describes the behavior of the graph of the function at the "ends" of the x-axis. In other words, the end behavior of a function describes the trend of the graph if we look to the right end of the x-axis (as x approaches +∞ ) and to the left end of the x-axis (as x approaches −∞ ).
