Tan x = b/a find value of sqrt a+b /a-b + sqrt a-b/ a+b answer is 2 cos x / sqrt cos 2x

To solve the expression involving \( \tan x = \frac{b}{a} \) and find the value of \( \frac{\sqrt{a+b}}{a-b} + \frac{\sqrt{a-b}}{a+b} \), we can approach it step by step, employing trigonometric identities and algebraic manipulation.

Understanding the Given Information

The equation \( \tan x = \frac{b}{a} \) relates the opposite side \( b \) and the adjacent side \( a \) of a right triangle. This tells us that we can express \( a \) and \( b \) in terms of the angle \( x \):

• Let \( a = k \cos x \)
• Let \( b = k \sin x \) for some positive constant \( k \)

Substituting into the Expression

Now, we can substitute \( a \) and \( b \) into the expression \( \frac{\sqrt{a+b}}{a-b} + \frac{\sqrt{a-b}}{a+b} \):

• Calculate \( a + b = k \cos x + k \sin x = k (\cos x + \sin x) \)
• Calculate \( a - b = k \cos x - k \sin x = k (\cos x - \sin x) \)

Calculating the Square Roots

Now we need to compute the square roots:

• \( \sqrt{a+b} = \sqrt{k (\cos x + \sin x)} \)
• \( \sqrt{a-b} = \sqrt{k (\cos x - \sin x)} \)

Simplifying the Expression

Substituting these back into our expression gives us:

\( \frac{\sqrt{k (\cos x + \sin x)}}{k (\cos x - \sin x)} + \frac{\sqrt{k (\cos x - \sin x)}}{k (\cos x + \sin x)} \)

This simplifies to:

\( \frac{\sqrt{\cos x + \sin x}}{\cos x - \sin x} + \frac{\sqrt{\cos x - \sin x}}{\cos x + \sin x} \)

Combining the Terms

Finding a common denominator, we can rewrite the expression as:

\( \frac{(\sqrt{\cos x + \sin x})^2 + (\sqrt{\cos x - \sin x})^2}{(\cos x - \sin x)(\cos x + \sin x)} \)

Now, squaring the square roots leads to:

\( \frac{(\cos x + \sin x) + (\cos x - \sin x)}{(\cos^2 x - \sin^2 x)} = \frac{2 \cos x}{\cos 2x} \)

Final Result

Thus, the result simplifies to:

\( \frac{2 \cos x}{\sqrt{\cos 2x}} \)

This shows how using trigonometric identities and algebraic manipulation can lead us to the desired result. Remember, understanding the relationships between the sides of a right triangle and the angles involved is key to unlocking these kinds of problems!