Verify that the trigonometric equation is an identity. 1+ secx 1-secx 1- secx 1+ secx -=-4 cot²x secx Which of the following statements establishes the identity? OA. 1+ secx 1-secx (1+ secx)²-(1-secx)²4 secx 1-secx 1+ secx (1-secx)(1+ secx) 1-sec²x OB. 1+ secx 1-secx (1+cscx-(1-cscx 1-secx 1 secx (1-cscx)(1+cscx) OC. 1+ secx 1-secx 1- secx 1+ secx OD. 1+ secx 1-secx 1-secx 1+ secx 4cscx 1-csC²x (sinx+1)-(sinx-1) 4 sinx (sinx-1)(sinx+1) ²x-1 (cosx+1-(cosx-12 4cosx (cosx-1)(cosx+1) cos²x-1 4 secx tan x 4cScx cotx -=-4 cotx secx 4 sin x
The Correct Answer and Explanation is :
The given trigonometric equation to verify as an identity is:
[
\frac{1 + \sec x}{1 – \sec x} \cdot \frac{1 – \sec x}{1 + \sec x} = -4 \cot^2 x \sec x
]
We are tasked with verifying this and determining which of the provided steps establishes the identity.
Step 1: Simplify the left-hand side
The left-hand side (LHS) is:
[
\frac{1 + \sec x}{1 – \sec x} \cdot \frac{1 – \sec x}{1 + \sec x}
]
Observe that multiplying these two fractions simplifies as:
[
\frac{1 + \sec x}{1 – \sec x} \cdot \frac{1 – \sec x}{1 + \sec x} = 1
]
Thus, the LHS simplifies to 1.
Step 2: Examine the right-hand side
The right-hand side (RHS) is:
[
-4 \cot^2 x \sec x
]
Step 3: Compare LHS and RHS
For the equation to be an identity, the LHS must equal the RHS. However, the LHS simplifies to 1, while the RHS simplifies to a trigonometric expression involving (\cot^2 x) and (\sec x). Clearly, these two sides are not equal for all (x). This means the given equation is not an identity.
Correct Answer
None of the provided options establish the given equation as an identity because the equation itself is incorrect. The simplified LHS and RHS are not equal, and the equation fails to hold universally.