Math  /  Trigonometry

QuestionStatement Rule (csc2x1)sec2x\left(\csc ^{2} x-1\right) \sec ^{2} x =cot2xsec2x=\cot ^{2} x \sec ^{2} x
Rule? =(cos2xsin2x)sec2x=\left(\frac{\cos ^{2} x}{\sin ^{2} x}\right) \sec ^{2} x Rule? =(cos2xsin2x)(1cos2x)=\left(\frac{\cos ^{2} x}{\sin ^{2} x}\right)\left(\frac{1}{\cos ^{2} x}\right) Rule? =1sin2x=\frac{1}{\sin ^{2} x} Rule? =csc2x=\csc ^{2} x Rule?

Studdy Solution

STEP 1

1. We are given a trigonometric identity to verify and simplify.
2. The goal is to show the equivalence of the given expression to csc2x\csc^2 x.
3. We will use trigonometric identities to simplify the expression step by step.

STEP 2

1. Simplify the expression (csc2x1)sec2x(\csc^2 x - 1) \sec^2 x using trigonometric identities.
2. Verify each transformation step to ensure correctness.

STEP 3

Start with the expression (csc2x1)sec2x(\csc^2 x - 1) \sec^2 x. Use the Pythagorean identity csc2x1=cot2x\csc^2 x - 1 = \cot^2 x.
csc2x1=cot2x \csc^2 x - 1 = \cot^2 x
Thus, the expression becomes:
cot2xsec2x \cot^2 x \sec^2 x

STEP 4

Express cot2x\cot^2 x and sec2x\sec^2 x in terms of sine and cosine:
cot2x=cos2xsin2x \cot^2 x = \frac{\cos^2 x}{\sin^2 x} sec2x=1cos2x \sec^2 x = \frac{1}{\cos^2 x}
Combine these to get:
(cos2xsin2x)sec2x=(cos2xsin2x)(1cos2x) \left(\frac{\cos^2 x}{\sin^2 x}\right) \sec^2 x = \left(\frac{\cos^2 x}{\sin^2 x}\right) \left(\frac{1}{\cos^2 x}\right)

STEP 5

Simplify the expression by canceling cos2x\cos^2 x:
(cos2xsin2x)(1cos2x)=1sin2x \left(\frac{\cos^2 x}{\sin^2 x}\right) \left(\frac{1}{\cos^2 x}\right) = \frac{1}{\sin^2 x}

STEP 6

Recognize that 1sin2x\frac{1}{\sin^2 x} is the definition of csc2x\csc^2 x:
1sin2x=csc2x \frac{1}{\sin^2 x} = \csc^2 x
The final expression is csc2x\csc^2 x, confirming the identity.

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