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CosecA/(1 SecA ) ( 1 SecA)/CosecA =2Cosec^3A(SecA-1) prove?
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CosecA/(1 SecA ) ( 1 SecA)/CosecA =2Cosec^3A(SecA-1) prove?
**Proving the Given Identity:**

To prove the given identity, we need to simplify the left-hand side (LHS) and the right-hand side (RHS) separately and show that they are equal.

**Simplifying the LHS:**

The LHS of the given identity is:
CosecA/(1 SecA)

We can simplify this expression by multiplying both the numerator and the denominator by SecA:

CosecA/(1 SecA) = (CosecA * SecA) / (1 SecA * SecA)

Using the reciprocal identities:
CosecA * SecA = 1

The expression becomes:
1 / (1 * Sec^2A)

Simplifying further:
1 / Sec^2A = Cos^2A

**Simplifying the RHS:**

The RHS of the given identity is:
(1 SecA)/(CosecA) = (1 / SecA) * (CosecA / 1)

Using the reciprocal identities:
1 / SecA = CosA

The expression becomes:
CosA * (CosecA / 1) = CosA * CosecA

Using the identity:
CosA * CosecA = 1

**Comparing LHS and RHS:**

We have simplified the LHS to Cos^2A and the RHS to 1. Now, we need to show that Cos^2A = 1.

Using the Pythagorean identity:
Sin^2A + Cos^2A = 1

Rearranging the equation:
Cos^2A = 1 - Sin^2A

Using the reciprocal identity:
SinA = 1 / CosecA

Substituting SinA in terms of CosecA:
Cos^2A = 1 - (1 / CosecA)^2

Simplifying:
Cos^2A = 1 - (1 / (1 / SinA))^2
Cos^2A = 1 - (1 / (1 / (1 / CosecA)))^2

Simplifying further:
Cos^2A = 1 - (1 * CosecA)^2
Cos^2A = 1 - Cosec^2A
Cos^2A = Sin^2A

Since Sin^2A = Cos^2A, we have proved that the LHS and RHS are equal.

Therefore, the given identity is true:
CosecA/(1 SecA) = (1 SecA)/(CosecA)
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CosecA/(1 SecA ) ( 1 SecA)/CosecA =2Cosec^3A(SecA-1) prove?
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