Trigonometric Identities

To solve the problem, we need to evaluate the argument (principal value) of the complex expression:

1. Expressing Complex Numbers in Polar Form:
We know:

2. Apply Exponent Rules:
Using the property:
So,

3. Divide the Expressions:

=

4. Compute the Argument:
We only need the angle (argument) part:
Now reduce the angle to its principal value in .

Divide by : remainder
So,

Since is a full rotation, the angle reduces to
But lies in the 4th quadrant, and we must bring it into range.

Final Answer:
The value of the argument is

Firstly, recognize the relationships between the hyperbolic secant and the hyperbolic tangent functions: Substituting in these equations: Now for , using its equivalent in terms of : The hyperbolic cosine inverse, , can be derived from the definition: Solving this equation for typically results in: Solving this quadratic in terms of , we get , so . Thus, . Subtracting the two results: But, correcting the error in the sign and simplifying: Hence, the correct result, considering simplifications and accurate transformations, should be .

We are given the sides , , and in triangle . Our goal is to find the value of .

Step 1: Apply the Law of Cosines to find angle . The Law of Cosines states: Substituting the given values: Thus, (since is obtuse).

Step 2: Use the Law of Sines to find angles and . The Law of Sines states: Substituting and , we get: Since , we have: Similarly:

Step 3: Find and . First, : Next, can be computed using the identity: Using the values of and calculated earlier, we find:

Step 4: Compute Now, compute the ratio: Thus: Thus, the correct answer is:

First, recognize that and are trigonometric inverses, and equating their differences to specific values can be simplified if we match these values to known angles. Given that and , we find: Now, solve for knowing that: And since implies . Using this in the expression : Using the tangent addition formula:

To evaluate the right-hand side, we start by recognizing the symmetry and identity relations between the given angles: Thus, we can simplify the expression by noting these values are equal due to their complementary nature and periodic properties of sine and cosine functions: Using the identity for and : Since and Given the equation is , solve for : Since is acute and , the solution for is , not as initially suggested.

To solve the equation , substitute with and with to obtain: Simplifying, we have: Let . The equation becomes: Using the quadratic formula, , where , , and : Thus, or , translating to or . For , we get: Since is an acute angle, we choose the positive value: