Prove that :
`sqrt((1+sinA)/(1-sinA))-sqrt((1-sinA)/(1+sinA))=2tanA`

Answer

To prove the equation:

\[ \sqrt{\frac{1+\sin A}{1-\sin A}} - \sqrt{\frac{1-\sin A}{1+\sin A}} = 2\tan A \]

we will follow these steps:

### Step 1: Combine the terms on the left side
We start by finding a common denominator for the two square root terms. The common denominator is \(\sqrt{(1-\sin A)(1+\sin A)}\).

\[ \text{Let } x = \sqrt{\frac{1+\sin A}{1-\sin A}} - \sqrt{\frac{1-\sin A}{1+\sin A}} \]

Rewriting this with a common denominator gives:

\[ x = \frac{\sqrt{(1+\sin A)(1+\sin A)} - \sqrt{(1-\sin A)(1-\sin A)}}{\sqrt{(1-\sin A)(1+\sin A)}} \]

### Step 2: Simplify the numerator
The numerator simplifies to:

\[ \sqrt{(1+\sin A)^2} - \sqrt{(1-\sin A)^2} = (1+\sin A) - (1-\sin A) = 2\sin A \]

So we can rewrite \(x\) as:

\[ x = \frac{2\sin A}{\sqrt{(1-\sin A)(1+\sin A)}} \]

### Step 3: Simplify the denominator
The denominator simplifies as follows:

\[ \sqrt{(1-\sin A)(1+\sin A)} = \sqrt{1 - \sin^2 A} = \sqrt{\cos^2 A} = \cos A \]

### Step 4: Substitute back into the equation
Now substituting back into our expression for \(x\):

\[ x = \frac{2\sin A}{\cos A} \]

### Step 5: Recognize the trigonometric identity
Using the identity \(\tan A = \frac{\sin A}{\cos A}\):

\[ x = 2\tan A \]

### Conclusion
Thus, we have shown that:

\[ \sqrt{\frac{1+\sin A}{1-\sin A}} - \sqrt{\frac{1-\sin A}{1+\sin A}} = 2\tan A \]

Hence, the equation is proved.

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