Find [√(1+sin2A)+√(1-sin2A)]/√(1+sin2A)-√(1-sin2A)Given |tan A|

To simplify the expression \(\frac{\sqrt{1 + \sin 2A} + \sqrt{1 - \sin 2A}}{\sqrt{1 + \sin 2A} - \sqrt{1 - \sin 2A}}\) given \(|\tan A|\), we can begin by using some trigonometric identities and algebraic manipulation. Let's break it down step-by-step.

Step 1: Rewrite \(\sin 2A\)

Recall the double angle identity for sine: \(\sin 2A = 2 \sin A \cos A\). This will be useful to express \(\sin 2A\) in a different form:

Substituting \(\sin 2A\)

In our expression, we can replace \(\sin 2A\) with \(2 \sin A \cos A\). However, first, let's simplify the components \(\sqrt{1 + \sin 2A}\) and \(\sqrt{1 - \sin 2A}\).

Step 2: Simplifying the square roots

We can express the square roots using the Pythagorean identity:

• \(\sqrt{1 + \sin 2A} = \sqrt{1 + 2 \sin A \cos A}\)
• \(\sqrt{1 - \sin 2A} = \sqrt{1 - 2 \sin A \cos A}\)

Using the identity

Now, applying the identity, we can also see that:

• \(1 + \sin 2A = \cos^2 A + \sin^2 A + \sin 2A = (\cos A + \sin A)^2\)
• \(1 - \sin 2A = \cos^2 A + \sin^2 A - \sin 2A = (\cos A - \sin A)^2\)

Step 3: Replacing the square roots

This allows us to rewrite our square roots:

• \(\sqrt{1 + \sin 2A} = \cos A + \sin A\)
• \(\sqrt{1 - \sin 2A} = \cos A - \sin A\)

Step 4: Plugging back into the expression

Substituting these back into our original expression gives:

\(\frac{(\cos A + \sin A) + (\cos A - \sin A)}{(\cos A + \sin A) - (\cos A - \sin A)}\)

Combining like terms

Now simplify the numerator and denominator:

• For the numerator: \((\cos A + \sin A) + (\cos A - \sin A) = 2\cos A\)
• For the denominator: \((\cos A + \sin A) - (\cos A - \sin A) = 2\sin A\)

Step 5: Final simplification

Thus, our expression simplifies to:

\(\frac{2\cos A}{2\sin A} = \frac{\cos A}{\sin A} = \cot A\)

Conclusion

So, the simplified form of the expression \(\frac{\sqrt{1 + \sin 2A} + \sqrt{1 - \sin 2A}}{\sqrt{1 + \sin 2A} - \sqrt{1 - \sin 2A}}\) is \(\cot A\). Given \(|\tan A|\), you can determine that \(|\cot A| = \frac{1}{|\tan A|}\).