1. Prove that sec x - tan x = 1/(sec x + tan x)
2. Explain clearly why 0 < sec x - tan x < 1 for values of x such that 0 < x < 1/2pi
I have done the first question, please help me with the second, thank you very much
Please help me with A level maths?
2018-01-08 9:17 pm
回答 (4)
2018-01-08 9:57 pm
2.
For 0 < x < π/2
0 < cos x < 1 and sin x > 0
Then,
0 < cos x < (1 + sin x)
Divided by (1 + sin x):
0/(1 + sin x) < cos x/(1 + sinx) < (1 + sin x)/(1 + sin x)
0 < 1/[(1 + sin x)/cos x] < 1
0 < 1/[(1/cos x) + (sin x/cos x)] < 1
0 < 1/(sec x + tan x) < 1
But 1/(sec x + tan x) = sec x - tan x
Then, 0 < sec x - tan x < 1
For 0 < x < π/2
0 < cos x < 1 and sin x > 0
Then,
0 < cos x < (1 + sin x)
Divided by (1 + sin x):
0/(1 + sin x) < cos x/(1 + sinx) < (1 + sin x)/(1 + sin x)
0 < 1/[(1 + sin x)/cos x] < 1
0 < 1/[(1/cos x) + (sin x/cos x)] < 1
0 < 1/(sec x + tan x) < 1
But 1/(sec x + tan x) = sec x - tan x
Then, 0 < sec x - tan x < 1
2018-01-08 10:25 pm
Right side
1/(sec x + tan x)
multiply and dvivide by (sec x - tan x)
= (sec x -tan x) /((sec x + tan x )(sec x - tan x))
= (sec x - tan x) / (sec^2 x - tan^2 x)
= (sec x - tan x) / (sec^2 x - (sec^2 x -1))
= (sec x - tan x) / 1
= sec x - tan x (left side)
1/(sec x + tan x)
multiply and dvivide by (sec x - tan x)
= (sec x -tan x) /((sec x + tan x )(sec x - tan x))
= (sec x - tan x) / (sec^2 x - tan^2 x)
= (sec x - tan x) / (sec^2 x - (sec^2 x -1))
= (sec x - tan x) / 1
= sec x - tan x (left side)
2018-01-08 9:56 pm
1/(sec???? + tan????) = 1/(1/cos???? + sin????/cos????)
= 1/((1+sin????)/cos????)
= cos????/(1+sin????)
= 1/((1+sin????)/cos????)
= 1/(1/cos???? + sin????/cos????)
= 1/(sec???? + tan????)
= 1/((1+sin????)/cos????)
= cos????/(1+sin????)
= 1/((1+sin????)/cos????)
= 1/(1/cos???? + sin????/cos????)
= 1/(sec???? + tan????)
收錄日期: 2021-04-18 18:07:47
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