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Helen [10]
3 years ago
7

Solve for the angle B tan 2B = cot 2B

Mathematics
1 answer:
Sedbober [7]3 years ago
3 0

\tan2B=\cot2B

\tan2B-\cot2B=0

\tan2B-\dfrac1{\tan2B}=0

\dfrac{\tan^22B-1}{\tan2B}=0

Note that we can't have \tan2B=0. Meanwhile,

\tan^22B-1=0\implies\tan^22B=1\implies\tan2B=\pm1

\tan2B=\pm1\implies2B=\pm\dfrac\pi4+n\pi

where n is any integer.

\implies\boxed{B=\pm\dfrac\pi8+\dfrac{n\pi}2}

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