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Tcecarenko [31]
3 years ago
14

Cot? 2x + cos2x + sin’ 2x = csc? 2x Need help with 40

Mathematics
1 answer:
kkurt [141]3 years ago
4 0

Answer:

\rule{100mm}{0.1mm}

    \cot^2(2x)+\cos^2(2x)+\sin^2(2x)=\csc^2(2x)

\dfrac{\cos^2(2x)}{\sin^2(2x)} +[(\cos^2(2x)+\sin^2(2x)]=\csc^2(2x)

                               \dfrac{\cos^2(2x)}{\sin^2(2x)} +1=\csc^2(2x)

                    \dfrac{\cos^2(2x)+\sin^2(2x)}{\sin^2(2x)}=\csc^2(2x)

                                    \dfrac{1}{sin^2(2x)} =\csc^2(2x)

                                     \boxed{\csc^2(2x)=\csc^2(2x)} ✔

\rule{100mm}{0.1mm}

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I'm just trying to make this as simple as possible for you.

The value of m is 5.

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We must follow a few simple methods to solve this problem.

Well, four to be precise.

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Then, we cancel the terms that are in both the numerator and denominator.

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Finally, we divide.

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