Asked by Arjun
If sec0-tan0=4
Then prove that cos0= 8/7
Then prove that cos0= 8/7
Answers
Answered by
Steve
Clearly you have a typo, since cosine is never greater than 1.
secθ-tanθ = 4
Since we want cosθ at the end, let's change stuff now:
1/cosθ - sinθ/cosθ = 4
(1-sinθ) = 4cosθ
1-2sinθ+sin^2θ = 16-16sin^2θ
17sin^2θ-2sinθ-15 = 0
(15sinθ-17)(sinθ+1) = 0
sin17sin^2θ-2sinθ-15 = 15/17 or -1
So, cosθ = 8/17
The other angle is a spurious solution
secθ-tanθ = 4
Since we want cosθ at the end, let's change stuff now:
1/cosθ - sinθ/cosθ = 4
(1-sinθ) = 4cosθ
1-2sinθ+sin^2θ = 16-16sin^2θ
17sin^2θ-2sinθ-15 = 0
(15sinθ-17)(sinθ+1) = 0
sin17sin^2θ-2sinθ-15 = 15/17 or -1
So, cosθ = 8/17
The other angle is a spurious solution
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