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fomenos
3 years ago
13

Octagon ABCDEFGH and its dilation, octagon A’B’C’D’E’F’G’H’, are shown on the coordinate plane below.

Mathematics
1 answer:
bazaltina [42]3 years ago
3 0

Answer: 2


Step-by-step explanation:


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If tanA+sinA=m and tanA-sinA=n.prove that m^2-n^2=4√mn
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Let a=\tan A and b=\sin A. Then

m^2-n^2=(a+b)^2-(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)=4ab

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and

mn=(a+b)(a-b)=a^2-b^2

\implies4\sqrt{mn}=4\sqrt{\tan^2A-\sin^2A}

The expression under the square root can be rewritten as

\tan^2A-\sin^2A=\dfrac{\sin^2A}{\cos^2A}-\sin^2A=\sin^2A\left(\dfrac1{\cos^2A}-1\right)=\sin^2A(\sec^2A-1)

Recall that

\sin^2A+\cos^2A=1\implies\tan^2A+1=\sec^2A

so that

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and assuming \sin A>0 and \tan A>0, we end up with

4\sqrt{\tan^2A-\sin^2A}=4\tan A\sin A

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as required.

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