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Math  /  Trigonometry

QuestionStatement Rule (csc⁡2x−1)sec⁡2x\left(\csc ^{2} x-1\right) \sec ^{2} x =cot⁡2xsec⁡2x=\cot ^{2} x \sec ^{2} x
Rule? =(cos⁡2xsin⁡2x)sec⁡2x=\left(\frac{\cos ^{2} x}{\sin ^{2} x}\right) \sec ^{2} x Rule? =(cos⁡2xsin⁡2x)(1cos⁡2x)=\left(\frac{\cos ^{2} x}{\sin ^{2} x}\right)\left(\frac{1}{\cos ^{2} x}\right) Rule? =1sin⁡2x=\frac{1}{\sin ^{2} x} Rule? =csc⁡2x=\csc ^{2} x Rule?

Studdy Solution
Recognize that 1sin⁡2x\frac{1}{\sin^2 x} is the definition of csc⁡2x\csc^2 x:
1sin⁡2x=csc⁡2x \frac{1}{\sin^2 x} = \csc^2 x
The final expression is csc⁡2x\csc^2 x, confirming the identity.

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