Trigonometric Functions

Tan 3θ = -1
Tan 3θ = - Tan = Tan (π - ) = Tan (2π - ) = Tan (3π - ) = Tan (4π - ) = Tan (5π - ) = Tan (6π - )
Tan 3θ = tan = tan = tan = tan = tan = tan
3θ = = = = = =
θ = = = = = =
So principal solutions are { , , , , , }
Therefore the correct option is (B).

We are tasked with solving the trigonometric equation. Let's break it down step by step.

Step 1: Start with the given equation:
The given equation is:

Step 2: Multiply through by :
To eliminate the denominator, we multiply both sides of the equation by :

Step 3: Use the identity :
We use the Pythagorean identity to express in terms of : Substituting this into the equation, we get:

Step 4: Simplify the equation:
Expanding the right side: Bringing all terms to one side: This is a quadratic equation in terms of .

Step 5: Solve using the quadratic formula:
The quadratic equation is: Applying the quadratic formula: For , we get: Simplifying: So we get two possible solutions:

Step 6: Find the values of :
For , we have: For , we have: These are the two solutions in the range .

Final Answer:
Therefore, the given equation has two solutions in the range , which corresponds to option (D).

We are tasked with finding the number of solutions to the equation in the interval . Let's analyze the behavior of the functions involved.

Step 1: Rewrite the equation using trigonometric identities:
The tangent and secant functions can be rewritten as:

Step 2: Combine the fractions on the left-hand side:
To combine the fractions, we get:

Step 3: Multiply both sides by :
Assuming , we multiply both sides by to eliminate the denominator:

Step 4: Use the Pythagorean identity:
Using the identity , we substitute into the equation:

Step 5: Rearrange the equation:
Expanding and simplifying: Bringing all terms to one side: This is a quadratic equation in terms of .

Step 6: Factor the quadratic equation:
We can factor the quadratic equation as: This gives two possible solutions:

Step 7: Solve for :
- For , in the interval , the solutions are: - For , in the interval , the solution is:

Step 8: Check for extraneous solutions:
Since we multiplied both sides of the equation by , we need to check if any solution results in , which would make and undefined.
- For , we calculate: and Thus, is a valid solution.
- For , we calculate: and Thus, is a valid solution.
- For , , so and are undefined. Therefore, is not a solution.

Final Answer:
Therefore, the solutions to the equation in the interval are and .