Answer:
(1-cos2A) /(1+cos2A) =tan²A
Proof:
We know that,
cos(A+B) =cosA.cosB-sinA.sinB
=>cos2A=cos(A+A)
=>cos2A=cosA.cosA - sinA.sinA
=>cos2A=cos²A-sin²A
=>cos2A=(cos²A-sin²A)/(cos²A+sin²A
Since {cos²A+sin²A=1}
Divide the numerator & the denominator by (cos²A) to get,
cos2A = {(cos²A-sin²A) ÷cos²A} / {(cos²A+sin²A) ÷cos²A}
cos2A ={(1-tan²A)/(1+tan²A)}
Then,
1-cos2A = 1-[{(1–tan²A)/(1+tan²A)}]
1-cos2A =(1+tan²A-1+tan²A)/(1+tan²A)
1-cos2A=(2tan²A)/(1+tan²A)
And now.......
1+cos2A=1+[{(1-tan²A)/(1+tan²A)}]
1+cos2A={1+tan²A+1-tan²A}/{1+tan²A}
1+cos2A=2/(1+tan²A)
So now,
(1-cos2A)/(1+cos2A)= {2tan²A/(1+tan²A)}÷{2/(1+tan²A)}
={(2tan²A)(1+tan²A)}÷{2(1+tan²A)}
=tan²A
Step-by-step explanation:
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Answer:
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Step-by-step explanation:
a respuesta an cuando (f 94)=94
First use distributive property on both sides of the equation.
-9(-16 - f) = 4(1 - 3f)
144 + 9f = 4 - 12f
Then isolate f on one side of the equation:
3f = -140
f = 46.67 --> (which is equal to -140/3)
I'm sorry I couldn't help more. I don't know why it's wrong. Maybe try putting in it decial form instead.
