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OpenStudy (anonymous):
The angle A is such that
secA + tanA = 2
Show that
secA-tanA=1/2
Hence, find the exact value of cosA.
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OpenStudy (anonymous):
use (secA + tanA )(secA-tanA)=1
thus secA-tanA=1/2
OpenStudy (anonymous):
Solve simultaneously
2secA=5/2
cosA=4/5
OpenStudy (anonymous):
Don't really understand... Can you start from the beginning?
OpenStudy (anonymous):
You get the first part? right
\[\sec^2A-\tan^2A=1=(secA-tanA)(secA+tanA)\]
OpenStudy (anonymous):
Not the "thus secA-tanA=1/2"
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OpenStudy (anonymous):
Use result from first part with what is already given and add the two
\[(secA+tanA)+(secA-tanA)=2+1/2\]
OpenStudy (anonymous):
\[\sec A + \tan A = \frac 1{(\sec A - \tan A)} \]
OpenStudy (anonymous):
Oh Ok.
OpenStudy (anonymous):
Note \(A \neq n\pi + \frac{\pi}{2} \) and \(n \in \mathbb{z} \)
OpenStudy (anonymous):
Ok, so how do you get the second part of the answer? @NotSObright and @FoolForMath
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OpenStudy (anonymous):
Add the two equation.
OpenStudy (anonymous):
Already mentioned
OpenStudy (anonymous):
order order think think :)
OpenStudy (anonymous):
But how did you get this in the first place?
2secA=5/2
cosA=4/5
I'm trying!... I'm trying!.... It's difficult for me though.. :/
OpenStudy (anonymous):
sec A = 1/ cos A
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OpenStudy (anonymous):
Still lost... :( My brain isn't working well today...
OpenStudy (anonymous):
|dw:1331036895119:dw|
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