Prove that :
[tex] \frac{secA}{secA + 1} + \frac{secA}{secA - 1} = 2cosec {}^{2} A[/tex]
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Prove that :
[tex] \frac{secA}{secA + 1} + \frac{secA}{secA - 1} = 2cosec {}^{2} A[/tex]
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There are two ways to solve this:
Here's the step-by-step solution:
Step 1: Write the given equation
```
secA/(secA + 1) + secA/(secA - 1) = 2cosec^2 A
```
Step 2: Simplify the left-hand side (LHS) of the equation
```
LHS = secA(secA - 1) + secA(secA + 1) / (sec^2 A - 1)
= 2sec^2 A / (sec^2 A - 1)
```
Step 3: Use the identity sec^2 A = 1 + tan^2 A
```
LHS = 2(1 + tan^2 A) / ((1 + tan^2 A) - 1)
= 2(1 + tan^2 A) / tan^2 A
```
Step 4: Use the identity tan^2 A = cot^-2 A - 1
```
LHS = 2(1 + cot^-2 A - 1) / (cot^-2 A - 1)
= 2cot^-2 A / cot^-2 A
= 2cosec^2 A
```
So, we have proved that the left-hand side is equal to the right-hand side of the equation. Hence, the given trigonometric identity is true. I hope this helps! Let me know if you have any other questions.